How to evaluate this limit: $lim_{xto 0}frac{sqrt{x+1}-1}{x} = frac12$?
How do I evaluate the limit of $$lim_{xto 0}frac{sqrt{x+1}-1}{x} = frac{1}{2}$$? As $x$ approaches $0$, I know the answer is $frac{1}{2}$, but I got this question wrong. I think you have to multiply by the conjugate of the numerator? Can someone explain the steps to this, I must be doing something wrong when I multiply.
calculus limits radicals
edited Sep 5 '16 at 18:40
Martin Sleziak
44.7k7115270
asked Sep 10 '13 at 17:46
kb95825
47114
add a comment |
How do I evaluate the limit of $$lim_{xto 0}frac{sqrt{x+1}-1}{x} = frac{1}{2}$$? As $x$ approaches $0$, I know the answer is $frac{1}{2}$, but I got this question wrong. I think you have to multiply by the conjugate of the numerator? Can someone explain the steps to this, I must be doing something wrong when I multiply.
calculus limits radicals
edited Sep 5 '16 at 18:40
Martin Sleziak
44.7k7115270
asked Sep 10 '13 at 17:46
kb95825
47114
2
Hint: Recall the definition of a derivative.
– Andrew D
Sep 10 '13 at 17:47
For some basic information about writing math at this site see e.g. here, here, here and here.
– Martin Sleziak
Sep 10 '13 at 17:50
1
The problem with referring to the limit definition of derivative is that this is a sort of problem that appears in an introductory course well before derivatives are defined. (Afterwards, naturally, it is clearer what can be done with this limit...)
– colormegone
Sep 10 '13 at 17:52
add a comment |
How do I evaluate the limit of $$lim_{xto 0}frac{sqrt{x+1}-1}{x} = frac{1}{2}$$? As $x$ approaches $0$, I know the answer is $frac{1}{2}$, but I got this question wrong. I think you have to multiply by the conjugate of the numerator? Can someone explain the steps to this, I must be doing something wrong when I multiply.
calculus limits radicals
edited Sep 5 '16 at 18:40
Martin Sleziak
44.7k7115270
asked Sep 10 '13 at 17:46
kb95825
47114
How do I evaluate the limit of $$lim_{xto 0}frac{sqrt{x+1}-1}{x} = frac{1}{2}$$? As $x$ approaches $0$, I know the answer is $frac{1}{2}$, but I got this question wrong. I think you have to multiply by the conjugate of the numerator? Can someone explain the steps to this, I must be doing something wrong when I multiply.
calculus limits radicals
calculus limits radicals
edited Sep 5 '16 at 18:40
Martin Sleziak
44.7k7115270
asked Sep 10 '13 at 17:46
kb95825
47114
edited Sep 5 '16 at 18:40
Martin Sleziak
44.7k7115270
asked Sep 10 '13 at 17:46
kb95825
47114
edited Sep 5 '16 at 18:40
Martin Sleziak
44.7k7115270
edited Sep 5 '16 at 18:40
Martin Sleziak
44.7k7115270
edited Sep 5 '16 at 18:40
Martin Sleziak
44.7k7115270
44.7k7115270
asked Sep 10 '13 at 17:46
kb95825
47114
asked Sep 10 '13 at 17:46
kb95825
47114
asked Sep 10 '13 at 17:46
kb95825
47114
47114
2
Hint: Recall the definition of a derivative.
– Andrew D
Sep 10 '13 at 17:47
For some basic information about writing math at this site see e.g. here, here, here and here.
– Martin Sleziak
Sep 10 '13 at 17:50
1
The problem with referring to the limit definition of derivative is that this is a sort of problem that appears in an introductory course well before derivatives are defined. (Afterwards, naturally, it is clearer what can be done with this limit...)
– colormegone
Sep 10 '13 at 17:52
add a comment |
2
Hint: Recall the definition of a derivative.
– Andrew D
Sep 10 '13 at 17:47
For some basic information about writing math at this site see e.g. here, here, here and here.
– Martin Sleziak
Sep 10 '13 at 17:50
1
The problem with referring to the limit definition of derivative is that this is a sort of problem that appears in an introductory course well before derivatives are defined. (Afterwards, naturally, it is clearer what can be done with this limit...)
– colormegone
Sep 10 '13 at 17:52
2
2
Hint: Recall the definition of a derivative.
– Andrew D
Sep 10 '13 at 17:47
Hint: Recall the definition of a derivative.
– Andrew D
Sep 10 '13 at 17:47
For some basic information about writing math at this site see e.g. here, here, here and here.
– Martin Sleziak
Sep 10 '13 at 17:50
For some basic information about writing math at this site see e.g. here, here, here and here.
– Martin Sleziak
Sep 10 '13 at 17:50
1
1
The problem with referring to the limit definition of derivative is that this is a sort of problem that appears in an introductory course well before derivatives are defined. (Afterwards, naturally, it is clearer what can be done with this limit...)
– colormegone
Sep 10 '13 at 17:52
The problem with referring to the limit definition of derivative is that this is a sort of problem that appears in an introductory course well before derivatives are defined. (Afterwards, naturally, it is clearer what can be done with this limit...)
– colormegone
Sep 10 '13 at 17:52
add a comment |
4 Answers
4
active
oldest
votes
$require{cancel}$
Multiply numerator and denominator by the conjugate of the numerator: $$sqrt{x+1} + 1$$ then evaluate the limit.
When we multiply by the conjugate, recall how we factor the difference of squares: $$(sqrt a - b) cdot (sqrt a + b) = (sqrt a)^2 - b^2 = a - b^2$$
$$dfrac {sqrt{x+1} - 1}{x} cdot frac{sqrt{x+1} + 1}{sqrt{x+1} + 1} = dfrac {(x + 1) - 1 }{x(sqrt{x+1} + 1)}= dfrac {cancel{x}}{cancel{x}(sqrt{x+1} + 1)} = dfrac 1{sqrt{x + 1} + 1}$$
Now we need only to evaluate $$lim_{x to 0} dfrac 1{sqrt{x + 1} + 1}$$
I trust you can do that.
answered Sep 10 '13 at 17:47
amWhy
191k28224439
what do you get once you multiply? I think I'm doing my math wrong when I multiply the numerator by the conjugate.
– kb95825
Sep 10 '13 at 17:50
Recall the difference of squares: the numerator becomes: $$(x+1) - 1 = x$$
– amWhy
Sep 10 '13 at 17:51
@amWhy I suggest you fix the braces in the remainders ;-)
– AlexR
Sep 10 '13 at 17:57
@kb95825 Recall that we need to multiply numerator and denominator by the conjugate of the numerator: that is like multiplying by $1$, so it doesn't change the original function.
– amWhy
Sep 10 '13 at 17:57
yes thank you :)
– kb95825
Sep 10 '13 at 21:25
|
show 1 more comment
Method 1 (basic)
$$frac{sqrt{x+1} - 1}{x} stackrel{sqrt{x+1}^2 = |x+1|}{=} frac{|x + 1| - 1}{x (sqrt{x+1} + 1)} stackrel{x+1 geq 0, text{ for well-def.}}{=} frac{1}{sqrt{x + 1} + 1} to frac{1}{1+1} = frac{1}{2}$$
as $xto 0$, because $xmapsto sqrt{x}$ is continuous (and so the limit can be "used as input")
Method 2 (derivative)
$$frac{sqrt{x+1} - 1}{x} = frac{f(1 + x) - f(1)}{x}$$
where $f(y) = sqrt{y}$, thus the limit is
$$f'(1) = frac{1}{2 sqrt{1}} = frac{1}{2}$$
Method 3 (l'Hospital)
Since the form is $frac{0}{0}$, l'Hospital can be applied and gives
$$lim_{xto 0} frac{sqrt{x+1} - 1}{x} = lim_{xto 0} frac{frac{1}{2sqrt{x+1}}}{1} = lim_{xto 0} frac{1}{2sqrt{x + 1}} = frac{1}{2 sqrt{1}} = frac{1}{2}$$
answered Sep 10 '13 at 17:51
AlexR
22.6k12349
Perhaps it is also worth mentioning in method 1 that $(sqrt{x+1})^2=|x+1|$, but since the expression in question is only defined for $x+1ge0$, we can replace $|x+1|$ by $(x+1)$.
– Martin Sleziak
Sep 10 '13 at 17:54
@MartinSleziak Yep, that's true. I'll add it
– AlexR
Sep 10 '13 at 17:54
add a comment |
Alternative solution: Write $ t= sqrt{x+1}$ then $x=t^2-1$ and we get $$lim_{xto 0}frac{sqrt{x+1}-1}{x} = lim_{tto 1}frac{t-1}{t^2-1} =lim_{tto 1}frac{t-1}{(t-1)(t+1)}=frac12$$
answered Dec 9 '18 at 20:42
greedoid
38k114794
add a comment |
$x>-1$.
$x= (x +1)- 1= sqrt{(x+1)}^2-1^2$
$=(sqrt{(x+1)}-1)(sqrt{x+1}+1)$.
$dfrac{sqrt{(x+1)}-1}{x} =dfrac{x}{(sqrt{x+1}+1)x}.$
Take the limit $x rightarrow 0$.
answered Dec 9 '18 at 21:46
Peter Szilas
10.7k2720
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f489699%2fhow-to-evaluate-this-limit-lim-x-to-0-frac-sqrtx1-1x-frac12%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$require{cancel}$
Multiply numerator and denominator by the conjugate of the numerator: $$sqrt{x+1} + 1$$ then evaluate the limit.
When we multiply by the conjugate, recall how we factor the difference of squares: $$(sqrt a - b) cdot (sqrt a + b) = (sqrt a)^2 - b^2 = a - b^2$$
$$dfrac {sqrt{x+1} - 1}{x} cdot frac{sqrt{x+1} + 1}{sqrt{x+1} + 1} = dfrac {(x + 1) - 1 }{x(sqrt{x+1} + 1)}= dfrac {cancel{x}}{cancel{x}(sqrt{x+1} + 1)} = dfrac 1{sqrt{x + 1} + 1}$$
Now we need only to evaluate $$lim_{x to 0} dfrac 1{sqrt{x + 1} + 1}$$
I trust you can do that.
answered Sep 10 '13 at 17:47
amWhy
191k28224439
what do you get once you multiply? I think I'm doing my math wrong when I multiply the numerator by the conjugate.
– kb95825
Sep 10 '13 at 17:50
Recall the difference of squares: the numerator becomes: $$(x+1) - 1 = x$$
– amWhy
Sep 10 '13 at 17:51
@amWhy I suggest you fix the braces in the remainders ;-)
– AlexR
Sep 10 '13 at 17:57
@kb95825 Recall that we need to multiply numerator and denominator by the conjugate of the numerator: that is like multiplying by $1$, so it doesn't change the original function.
– amWhy
Sep 10 '13 at 17:57
yes thank you :)
– kb95825
Sep 10 '13 at 21:25
|
show 1 more comment
$require{cancel}$
Multiply numerator and denominator by the conjugate of the numerator: $$sqrt{x+1} + 1$$ then evaluate the limit.
When we multiply by the conjugate, recall how we factor the difference of squares: $$(sqrt a - b) cdot (sqrt a + b) = (sqrt a)^2 - b^2 = a - b^2$$
$$dfrac {sqrt{x+1} - 1}{x} cdot frac{sqrt{x+1} + 1}{sqrt{x+1} + 1} = dfrac {(x + 1) - 1 }{x(sqrt{x+1} + 1)}= dfrac {cancel{x}}{cancel{x}(sqrt{x+1} + 1)} = dfrac 1{sqrt{x + 1} + 1}$$
Now we need only to evaluate $$lim_{x to 0} dfrac 1{sqrt{x + 1} + 1}$$
I trust you can do that.
answered Sep 10 '13 at 17:47
amWhy
191k28224439
what do you get once you multiply? I think I'm doing my math wrong when I multiply the numerator by the conjugate.
– kb95825
Sep 10 '13 at 17:50
Recall the difference of squares: the numerator becomes: $$(x+1) - 1 = x$$
– amWhy
Sep 10 '13 at 17:51
@amWhy I suggest you fix the braces in the remainders ;-)
– AlexR
Sep 10 '13 at 17:57
@kb95825 Recall that we need to multiply numerator and denominator by the conjugate of the numerator: that is like multiplying by $1$, so it doesn't change the original function.
– amWhy
Sep 10 '13 at 17:57
yes thank you :)
– kb95825
Sep 10 '13 at 21:25
|
show 1 more comment
$require{cancel}$
Multiply numerator and denominator by the conjugate of the numerator: $$sqrt{x+1} + 1$$ then evaluate the limit.
When we multiply by the conjugate, recall how we factor the difference of squares: $$(sqrt a - b) cdot (sqrt a + b) = (sqrt a)^2 - b^2 = a - b^2$$
$$dfrac {sqrt{x+1} - 1}{x} cdot frac{sqrt{x+1} + 1}{sqrt{x+1} + 1} = dfrac {(x + 1) - 1 }{x(sqrt{x+1} + 1)}= dfrac {cancel{x}}{cancel{x}(sqrt{x+1} + 1)} = dfrac 1{sqrt{x + 1} + 1}$$
Now we need only to evaluate $$lim_{x to 0} dfrac 1{sqrt{x + 1} + 1}$$
I trust you can do that.
answered Sep 10 '13 at 17:47
amWhy
191k28224439
$require{cancel}$
Multiply numerator and denominator by the conjugate of the numerator: $$sqrt{x+1} + 1$$ then evaluate the limit.
When we multiply by the conjugate, recall how we factor the difference of squares: $$(sqrt a - b) cdot (sqrt a + b) = (sqrt a)^2 - b^2 = a - b^2$$
$$dfrac {sqrt{x+1} - 1}{x} cdot frac{sqrt{x+1} + 1}{sqrt{x+1} + 1} = dfrac {(x + 1) - 1 }{x(sqrt{x+1} + 1)}= dfrac {cancel{x}}{cancel{x}(sqrt{x+1} + 1)} = dfrac 1{sqrt{x + 1} + 1}$$
Now we need only to evaluate $$lim_{x to 0} dfrac 1{sqrt{x + 1} + 1}$$
I trust you can do that.
answered Sep 10 '13 at 17:47
amWhy
191k28224439
edited Dec 9 '18 at 20:39
answered Sep 10 '13 at 17:47
amWhy
191k28224439
answered Sep 10 '13 at 17:47
amWhy
191k28224439
answered Sep 10 '13 at 17:47
amWhy
191k28224439
191k28224439
what do you get once you multiply? I think I'm doing my math wrong when I multiply the numerator by the conjugate.
– kb95825
Sep 10 '13 at 17:50
Recall the difference of squares: the numerator becomes: $$(x+1) - 1 = x$$
– amWhy
Sep 10 '13 at 17:51
@amWhy I suggest you fix the braces in the remainders ;-)
– AlexR
Sep 10 '13 at 17:57
@kb95825 Recall that we need to multiply numerator and denominator by the conjugate of the numerator: that is like multiplying by $1$, so it doesn't change the original function.
– amWhy
Sep 10 '13 at 17:57
yes thank you :)
– kb95825
Sep 10 '13 at 21:25
|
show 1 more comment
what do you get once you multiply? I think I'm doing my math wrong when I multiply the numerator by the conjugate.
– kb95825
Sep 10 '13 at 17:50
Recall the difference of squares: the numerator becomes: $$(x+1) - 1 = x$$
– amWhy
Sep 10 '13 at 17:51
@amWhy I suggest you fix the braces in the remainders ;-)
– AlexR
Sep 10 '13 at 17:57
@kb95825 Recall that we need to multiply numerator and denominator by the conjugate of the numerator: that is like multiplying by $1$, so it doesn't change the original function.
– amWhy
Sep 10 '13 at 17:57
yes thank you :)
– kb95825
Sep 10 '13 at 21:25
what do you get once you multiply? I think I'm doing my math wrong when I multiply the numerator by the conjugate.
– kb95825
Sep 10 '13 at 17:50
what do you get once you multiply? I think I'm doing my math wrong when I multiply the numerator by the conjugate.
– kb95825
Sep 10 '13 at 17:50
Recall the difference of squares: the numerator becomes: $$(x+1) - 1 = x$$
– amWhy
Sep 10 '13 at 17:51
Recall the difference of squares: the numerator becomes: $$(x+1) - 1 = x$$
– amWhy
Sep 10 '13 at 17:51
@amWhy I suggest you fix the braces in the remainders ;-)
– AlexR
Sep 10 '13 at 17:57
@amWhy I suggest you fix the braces in the remainders ;-)
– AlexR
Sep 10 '13 at 17:57
@kb95825 Recall that we need to multiply numerator and denominator by the conjugate of the numerator: that is like multiplying by $1$, so it doesn't change the original function.
– amWhy
Sep 10 '13 at 17:57
@kb95825 Recall that we need to multiply numerator and denominator by the conjugate of the numerator: that is like multiplying by $1$, so it doesn't change the original function.
– amWhy
Sep 10 '13 at 17:57
yes thank you :)
– kb95825
Sep 10 '13 at 21:25
yes thank you :)
– kb95825
Sep 10 '13 at 21:25
|
show 1 more comment
Method 1 (basic)
$$frac{sqrt{x+1} - 1}{x} stackrel{sqrt{x+1}^2 = |x+1|}{=} frac{|x + 1| - 1}{x (sqrt{x+1} + 1)} stackrel{x+1 geq 0, text{ for well-def.}}{=} frac{1}{sqrt{x + 1} + 1} to frac{1}{1+1} = frac{1}{2}$$
as $xto 0$, because $xmapsto sqrt{x}$ is continuous (and so the limit can be "used as input")
Method 2 (derivative)
$$frac{sqrt{x+1} - 1}{x} = frac{f(1 + x) - f(1)}{x}$$
where $f(y) = sqrt{y}$, thus the limit is
$$f'(1) = frac{1}{2 sqrt{1}} = frac{1}{2}$$
Method 3 (l'Hospital)
Since the form is $frac{0}{0}$, l'Hospital can be applied and gives
$$lim_{xto 0} frac{sqrt{x+1} - 1}{x} = lim_{xto 0} frac{frac{1}{2sqrt{x+1}}}{1} = lim_{xto 0} frac{1}{2sqrt{x + 1}} = frac{1}{2 sqrt{1}} = frac{1}{2}$$
answered Sep 10 '13 at 17:51
AlexR
22.6k12349
Perhaps it is also worth mentioning in method 1 that $(sqrt{x+1})^2=|x+1|$, but since the expression in question is only defined for $x+1ge0$, we can replace $|x+1|$ by $(x+1)$.
– Martin Sleziak
Sep 10 '13 at 17:54
@MartinSleziak Yep, that's true. I'll add it
– AlexR
Sep 10 '13 at 17:54
add a comment |
Method 1 (basic)
$$frac{sqrt{x+1} - 1}{x} stackrel{sqrt{x+1}^2 = |x+1|}{=} frac{|x + 1| - 1}{x (sqrt{x+1} + 1)} stackrel{x+1 geq 0, text{ for well-def.}}{=} frac{1}{sqrt{x + 1} + 1} to frac{1}{1+1} = frac{1}{2}$$
as $xto 0$, because $xmapsto sqrt{x}$ is continuous (and so the limit can be "used as input")
Method 2 (derivative)
$$frac{sqrt{x+1} - 1}{x} = frac{f(1 + x) - f(1)}{x}$$
where $f(y) = sqrt{y}$, thus the limit is
$$f'(1) = frac{1}{2 sqrt{1}} = frac{1}{2}$$
Method 3 (l'Hospital)
Since the form is $frac{0}{0}$, l'Hospital can be applied and gives
$$lim_{xto 0} frac{sqrt{x+1} - 1}{x} = lim_{xto 0} frac{frac{1}{2sqrt{x+1}}}{1} = lim_{xto 0} frac{1}{2sqrt{x + 1}} = frac{1}{2 sqrt{1}} = frac{1}{2}$$
answered Sep 10 '13 at 17:51
AlexR
22.6k12349
Perhaps it is also worth mentioning in method 1 that $(sqrt{x+1})^2=|x+1|$, but since the expression in question is only defined for $x+1ge0$, we can replace $|x+1|$ by $(x+1)$.
– Martin Sleziak
Sep 10 '13 at 17:54
@MartinSleziak Yep, that's true. I'll add it
– AlexR
Sep 10 '13 at 17:54
add a comment |
Method 1 (basic)
$$frac{sqrt{x+1} - 1}{x} stackrel{sqrt{x+1}^2 = |x+1|}{=} frac{|x + 1| - 1}{x (sqrt{x+1} + 1)} stackrel{x+1 geq 0, text{ for well-def.}}{=} frac{1}{sqrt{x + 1} + 1} to frac{1}{1+1} = frac{1}{2}$$
as $xto 0$, because $xmapsto sqrt{x}$ is continuous (and so the limit can be "used as input")
Method 2 (derivative)
$$frac{sqrt{x+1} - 1}{x} = frac{f(1 + x) - f(1)}{x}$$
where $f(y) = sqrt{y}$, thus the limit is
$$f'(1) = frac{1}{2 sqrt{1}} = frac{1}{2}$$
Method 3 (l'Hospital)
Since the form is $frac{0}{0}$, l'Hospital can be applied and gives
$$lim_{xto 0} frac{sqrt{x+1} - 1}{x} = lim_{xto 0} frac{frac{1}{2sqrt{x+1}}}{1} = lim_{xto 0} frac{1}{2sqrt{x + 1}} = frac{1}{2 sqrt{1}} = frac{1}{2}$$
answered Sep 10 '13 at 17:51
AlexR
22.6k12349
Method 1 (basic)
$$frac{sqrt{x+1} - 1}{x} stackrel{sqrt{x+1}^2 = |x+1|}{=} frac{|x + 1| - 1}{x (sqrt{x+1} + 1)} stackrel{x+1 geq 0, text{ for well-def.}}{=} frac{1}{sqrt{x + 1} + 1} to frac{1}{1+1} = frac{1}{2}$$
as $xto 0$, because $xmapsto sqrt{x}$ is continuous (and so the limit can be "used as input")
Method 2 (derivative)
$$frac{sqrt{x+1} - 1}{x} = frac{f(1 + x) - f(1)}{x}$$
where $f(y) = sqrt{y}$, thus the limit is
$$f'(1) = frac{1}{2 sqrt{1}} = frac{1}{2}$$
Method 3 (l'Hospital)
Since the form is $frac{0}{0}$, l'Hospital can be applied and gives
$$lim_{xto 0} frac{sqrt{x+1} - 1}{x} = lim_{xto 0} frac{frac{1}{2sqrt{x+1}}}{1} = lim_{xto 0} frac{1}{2sqrt{x + 1}} = frac{1}{2 sqrt{1}} = frac{1}{2}$$
answered Sep 10 '13 at 17:51
AlexR
22.6k12349
edited Sep 10 '13 at 17:58
answered Sep 10 '13 at 17:51
AlexR
22.6k12349
answered Sep 10 '13 at 17:51
AlexR
22.6k12349
answered Sep 10 '13 at 17:51
AlexR
22.6k12349
22.6k12349
Perhaps it is also worth mentioning in method 1 that $(sqrt{x+1})^2=|x+1|$, but since the expression in question is only defined for $x+1ge0$, we can replace $|x+1|$ by $(x+1)$.
– Martin Sleziak
Sep 10 '13 at 17:54
@MartinSleziak Yep, that's true. I'll add it
– AlexR
Sep 10 '13 at 17:54
add a comment |
Perhaps it is also worth mentioning in method 1 that $(sqrt{x+1})^2=|x+1|$, but since the expression in question is only defined for $x+1ge0$, we can replace $|x+1|$ by $(x+1)$.
– Martin Sleziak
Sep 10 '13 at 17:54
@MartinSleziak Yep, that's true. I'll add it
– AlexR
Sep 10 '13 at 17:54
Perhaps it is also worth mentioning in method 1 that $(sqrt{x+1})^2=|x+1|$, but since the expression in question is only defined for $x+1ge0$, we can replace $|x+1|$ by $(x+1)$.
– Martin Sleziak
Sep 10 '13 at 17:54
Perhaps it is also worth mentioning in method 1 that $(sqrt{x+1})^2=|x+1|$, but since the expression in question is only defined for $x+1ge0$, we can replace $|x+1|$ by $(x+1)$.
– Martin Sleziak
Sep 10 '13 at 17:54
@MartinSleziak Yep, that's true. I'll add it
– AlexR
Sep 10 '13 at 17:54
@MartinSleziak Yep, that's true. I'll add it
– AlexR
Sep 10 '13 at 17:54
add a comment |
Alternative solution: Write $ t= sqrt{x+1}$ then $x=t^2-1$ and we get $$lim_{xto 0}frac{sqrt{x+1}-1}{x} = lim_{tto 1}frac{t-1}{t^2-1} =lim_{tto 1}frac{t-1}{(t-1)(t+1)}=frac12$$
answered Dec 9 '18 at 20:42
greedoid
38k114794
add a comment |
Alternative solution: Write $ t= sqrt{x+1}$ then $x=t^2-1$ and we get $$lim_{xto 0}frac{sqrt{x+1}-1}{x} = lim_{tto 1}frac{t-1}{t^2-1} =lim_{tto 1}frac{t-1}{(t-1)(t+1)}=frac12$$
answered Dec 9 '18 at 20:42
greedoid
38k114794
add a comment |
Alternative solution: Write $ t= sqrt{x+1}$ then $x=t^2-1$ and we get $$lim_{xto 0}frac{sqrt{x+1}-1}{x} = lim_{tto 1}frac{t-1}{t^2-1} =lim_{tto 1}frac{t-1}{(t-1)(t+1)}=frac12$$
answered Dec 9 '18 at 20:42
greedoid
38k114794
Alternative solution: Write $ t= sqrt{x+1}$ then $x=t^2-1$ and we get $$lim_{xto 0}frac{sqrt{x+1}-1}{x} = lim_{tto 1}frac{t-1}{t^2-1} =lim_{tto 1}frac{t-1}{(t-1)(t+1)}=frac12$$
answered Dec 9 '18 at 20:42
greedoid
38k114794
edited Dec 9 '18 at 21:36
answered Dec 9 '18 at 20:42
greedoid
38k114794
answered Dec 9 '18 at 20:42
greedoid
38k114794
answered Dec 9 '18 at 20:42
greedoid
38k114794
38k114794
add a comment |
add a comment |
$x>-1$.
$x= (x +1)- 1= sqrt{(x+1)}^2-1^2$
$=(sqrt{(x+1)}-1)(sqrt{x+1}+1)$.
$dfrac{sqrt{(x+1)}-1}{x} =dfrac{x}{(sqrt{x+1}+1)x}.$
Take the limit $x rightarrow 0$.
answered Dec 9 '18 at 21:46
Peter Szilas
10.7k2720
add a comment |
$x>-1$.
$x= (x +1)- 1= sqrt{(x+1)}^2-1^2$
$=(sqrt{(x+1)}-1)(sqrt{x+1}+1)$.
$dfrac{sqrt{(x+1)}-1}{x} =dfrac{x}{(sqrt{x+1}+1)x}.$
Take the limit $x rightarrow 0$.
answered Dec 9 '18 at 21:46
Peter Szilas
10.7k2720
add a comment |
$x>-1$.
$x= (x +1)- 1= sqrt{(x+1)}^2-1^2$
$=(sqrt{(x+1)}-1)(sqrt{x+1}+1)$.
$dfrac{sqrt{(x+1)}-1}{x} =dfrac{x}{(sqrt{x+1}+1)x}.$
Take the limit $x rightarrow 0$.
answered Dec 9 '18 at 21:46
Peter Szilas
10.7k2720
$x>-1$.
$x= (x +1)- 1= sqrt{(x+1)}^2-1^2$
$=(sqrt{(x+1)}-1)(sqrt{x+1}+1)$.
$dfrac{sqrt{(x+1)}-1}{x} =dfrac{x}{(sqrt{x+1}+1)x}.$
Take the limit $x rightarrow 0$.
answered Dec 9 '18 at 21:46
Peter Szilas
10.7k2720
answered Dec 9 '18 at 21:46
Peter Szilas
10.7k2720
answered Dec 9 '18 at 21:46
Peter Szilas
10.7k2720
answered Dec 9 '18 at 21:46
Peter Szilas
10.7k2720
10.7k2720
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f489699%2fhow-to-evaluate-this-limit-lim-x-to-0-frac-sqrtx1-1x-frac12%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
Hint: Recall the definition of a derivative.
– Andrew D
Sep 10 '13 at 17:47
For some basic information about writing math at this site see e.g. here, here, here and here.
– Martin Sleziak
Sep 10 '13 at 17:50
1
The problem with referring to the limit definition of derivative is that this is a sort of problem that appears in an introductory course well before derivatives are defined. (Afterwards, naturally, it is clearer what can be done with this limit...)
– colormegone
Sep 10 '13 at 17:52