Compute this limit $lim_{xto0}frac{sin(x^2+frac{1}{x})-sinfrac{1}{x}}{x}$ using L'Hôpital's rule
I have asked this problem before, but I can't understand the explanation, I couldn't understand how the sin multiply for cos, and too multiply for A + and - B: $$sin(A)-sin(B)=2sinleft(frac{A-B}{2}right)cosleft(frac{A+B}{2}right)$$ and I don't understand in this step how/why the $A-B$ and $A+B$ was replaced by $frac{x^2}{2}$ and $frac{x^2}{2}+frac{1}{x}$ :
$$lim_{xto0}frac{sinleft(x^2+frac1xright)-sinleft(frac1xright)}{x}= lim_{xto0}frac{2sinleft(frac{x^2}{2}right)cosleft(frac{x^2}{2}+frac1xright)}{x}.$$
calculus limits
edited Dec 12 '18 at 14:51
Lorenzo B.
1,8402520
asked Jun 19 '13 at 17:59
user73276user73276
390310
add a comment |
I have asked this problem before, but I can't understand the explanation, I couldn't understand how the sin multiply for cos, and too multiply for A + and - B: $$sin(A)-sin(B)=2sinleft(frac{A-B}{2}right)cosleft(frac{A+B}{2}right)$$ and I don't understand in this step how/why the $A-B$ and $A+B$ was replaced by $frac{x^2}{2}$ and $frac{x^2}{2}+frac{1}{x}$ :
$$lim_{xto0}frac{sinleft(x^2+frac1xright)-sinleft(frac1xright)}{x}= lim_{xto0}frac{2sinleft(frac{x^2}{2}right)cosleft(frac{x^2}{2}+frac1xright)}{x}.$$
calculus limits
edited Dec 12 '18 at 14:51
Lorenzo B.
1,8402520
asked Jun 19 '13 at 17:59
user73276user73276
390310
If you have asked this problem before, maybe you could give the link to that previous question?
– celtschk
Jun 19 '13 at 18:18
@user73276:i cant use L'hôpital's rule because $lim_{xto0}{sin(x^2+frac{1}{x})-sinfrac{1}{x}}neq0$
– M.H
Jun 19 '13 at 18:45
add a comment |
I have asked this problem before, but I can't understand the explanation, I couldn't understand how the sin multiply for cos, and too multiply for A + and - B: $$sin(A)-sin(B)=2sinleft(frac{A-B}{2}right)cosleft(frac{A+B}{2}right)$$ and I don't understand in this step how/why the $A-B$ and $A+B$ was replaced by $frac{x^2}{2}$ and $frac{x^2}{2}+frac{1}{x}$ :
$$lim_{xto0}frac{sinleft(x^2+frac1xright)-sinleft(frac1xright)}{x}= lim_{xto0}frac{2sinleft(frac{x^2}{2}right)cosleft(frac{x^2}{2}+frac1xright)}{x}.$$
calculus limits
edited Dec 12 '18 at 14:51
Lorenzo B.
1,8402520
asked Jun 19 '13 at 17:59
user73276user73276
390310
I have asked this problem before, but I can't understand the explanation, I couldn't understand how the sin multiply for cos, and too multiply for A + and - B: $$sin(A)-sin(B)=2sinleft(frac{A-B}{2}right)cosleft(frac{A+B}{2}right)$$ and I don't understand in this step how/why the $A-B$ and $A+B$ was replaced by $frac{x^2}{2}$ and $frac{x^2}{2}+frac{1}{x}$ :
$$lim_{xto0}frac{sinleft(x^2+frac1xright)-sinleft(frac1xright)}{x}= lim_{xto0}frac{2sinleft(frac{x^2}{2}right)cosleft(frac{x^2}{2}+frac1xright)}{x}.$$
calculus limits
calculus limits
edited Dec 12 '18 at 14:51
Lorenzo B.
1,8402520
asked Jun 19 '13 at 17:59
user73276user73276
390310
edited Dec 12 '18 at 14:51
Lorenzo B.
1,8402520
asked Jun 19 '13 at 17:59
user73276user73276
390310
edited Dec 12 '18 at 14:51
Lorenzo B.
1,8402520
edited Dec 12 '18 at 14:51
Lorenzo B.
1,8402520
edited Dec 12 '18 at 14:51
Lorenzo B.
1,8402520
1,8402520
asked Jun 19 '13 at 17:59
user73276user73276
390310
asked Jun 19 '13 at 17:59
user73276user73276
390310
asked Jun 19 '13 at 17:59
user73276user73276
390310
390310
If you have asked this problem before, maybe you could give the link to that previous question?
– celtschk
Jun 19 '13 at 18:18
@user73276:i cant use L'hôpital's rule because $lim_{xto0}{sin(x^2+frac{1}{x})-sinfrac{1}{x}}neq0$
– M.H
Jun 19 '13 at 18:45
add a comment |
If you have asked this problem before, maybe you could give the link to that previous question?
– celtschk
Jun 19 '13 at 18:18
@user73276:i cant use L'hôpital's rule because $lim_{xto0}{sin(x^2+frac{1}{x})-sinfrac{1}{x}}neq0$
– M.H
Jun 19 '13 at 18:45
If you have asked this problem before, maybe you could give the link to that previous question?
– celtschk
Jun 19 '13 at 18:18
If you have asked this problem before, maybe you could give the link to that previous question?
– celtschk
Jun 19 '13 at 18:18
@user73276:i cant use L'hôpital's rule because $lim_{xto0}{sin(x^2+frac{1}{x})-sinfrac{1}{x}}neq0$
– M.H
Jun 19 '13 at 18:45
@user73276:i cant use L'hôpital's rule because $lim_{xto0}{sin(x^2+frac{1}{x})-sinfrac{1}{x}}neq0$
– M.H
Jun 19 '13 at 18:45
add a comment |
2 Answers
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Hint:its easy to prove $$sin(x+y)+sin(x-y)=2sin(x)cos(y)$$
then put $y=frac{A+B}{2}$,$x=frac{A-B}{2}$
$$ sin(x)sim x$$ $$ cosxsim1-frac{x^2}{2}$$ because $$lim_{xto0}frac{sinx}{x}=lim_{xto0}frac{cosx}{1-frac{x^2}{2}}=1$$
answered Jun 19 '13 at 18:27
M.HM.H
7,2371553
add a comment |
It's easier if you use the plain old addition formula:
$$sin{left(x^2+frac{1}{x}right)} = cos{x^2}, sin{frac{1}{x}} + sin{x^2} ,cos{frac{1}{x}}$$
which behaves as
$$left(1-frac12 x^2right) sin{frac{1}{x}} + x^2 cos{frac{1}{x}}$$
so that
$$frac{sin{left(x^2+frac{1}{x}right)} - sin{frac{1}{x}}}{x} sim frac{x^2 cos{frac{1}{x}} - (1/2)x^2 sin{frac{1}{x}}}{x}$$
which goes to zero as $x to 0$.
answered Jun 19 '13 at 19:07
Ron GordonRon Gordon
122k14154263
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
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votes
Hint:its easy to prove $$sin(x+y)+sin(x-y)=2sin(x)cos(y)$$
then put $y=frac{A+B}{2}$,$x=frac{A-B}{2}$
$$ sin(x)sim x$$ $$ cosxsim1-frac{x^2}{2}$$ because $$lim_{xto0}frac{sinx}{x}=lim_{xto0}frac{cosx}{1-frac{x^2}{2}}=1$$
answered Jun 19 '13 at 18:27
M.HM.H
7,2371553
add a comment |
Hint:its easy to prove $$sin(x+y)+sin(x-y)=2sin(x)cos(y)$$
then put $y=frac{A+B}{2}$,$x=frac{A-B}{2}$
$$ sin(x)sim x$$ $$ cosxsim1-frac{x^2}{2}$$ because $$lim_{xto0}frac{sinx}{x}=lim_{xto0}frac{cosx}{1-frac{x^2}{2}}=1$$
answered Jun 19 '13 at 18:27
M.HM.H
7,2371553
add a comment |
Hint:its easy to prove $$sin(x+y)+sin(x-y)=2sin(x)cos(y)$$
then put $y=frac{A+B}{2}$,$x=frac{A-B}{2}$
$$ sin(x)sim x$$ $$ cosxsim1-frac{x^2}{2}$$ because $$lim_{xto0}frac{sinx}{x}=lim_{xto0}frac{cosx}{1-frac{x^2}{2}}=1$$
answered Jun 19 '13 at 18:27
M.HM.H
7,2371553
Hint:its easy to prove $$sin(x+y)+sin(x-y)=2sin(x)cos(y)$$
then put $y=frac{A+B}{2}$,$x=frac{A-B}{2}$
$$ sin(x)sim x$$ $$ cosxsim1-frac{x^2}{2}$$ because $$lim_{xto0}frac{sinx}{x}=lim_{xto0}frac{cosx}{1-frac{x^2}{2}}=1$$
answered Jun 19 '13 at 18:27
M.HM.H
7,2371553
edited Jun 19 '13 at 18:39
answered Jun 19 '13 at 18:27
M.HM.H
7,2371553
answered Jun 19 '13 at 18:27
M.HM.H
7,2371553
answered Jun 19 '13 at 18:27
M.HM.H
7,2371553
7,2371553
add a comment |
add a comment |
It's easier if you use the plain old addition formula:
$$sin{left(x^2+frac{1}{x}right)} = cos{x^2}, sin{frac{1}{x}} + sin{x^2} ,cos{frac{1}{x}}$$
which behaves as
$$left(1-frac12 x^2right) sin{frac{1}{x}} + x^2 cos{frac{1}{x}}$$
so that
$$frac{sin{left(x^2+frac{1}{x}right)} - sin{frac{1}{x}}}{x} sim frac{x^2 cos{frac{1}{x}} - (1/2)x^2 sin{frac{1}{x}}}{x}$$
which goes to zero as $x to 0$.
answered Jun 19 '13 at 19:07
Ron GordonRon Gordon
122k14154263
add a comment |
It's easier if you use the plain old addition formula:
$$sin{left(x^2+frac{1}{x}right)} = cos{x^2}, sin{frac{1}{x}} + sin{x^2} ,cos{frac{1}{x}}$$
which behaves as
$$left(1-frac12 x^2right) sin{frac{1}{x}} + x^2 cos{frac{1}{x}}$$
so that
$$frac{sin{left(x^2+frac{1}{x}right)} - sin{frac{1}{x}}}{x} sim frac{x^2 cos{frac{1}{x}} - (1/2)x^2 sin{frac{1}{x}}}{x}$$
which goes to zero as $x to 0$.
answered Jun 19 '13 at 19:07
Ron GordonRon Gordon
122k14154263
add a comment |
It's easier if you use the plain old addition formula:
$$sin{left(x^2+frac{1}{x}right)} = cos{x^2}, sin{frac{1}{x}} + sin{x^2} ,cos{frac{1}{x}}$$
which behaves as
$$left(1-frac12 x^2right) sin{frac{1}{x}} + x^2 cos{frac{1}{x}}$$
so that
$$frac{sin{left(x^2+frac{1}{x}right)} - sin{frac{1}{x}}}{x} sim frac{x^2 cos{frac{1}{x}} - (1/2)x^2 sin{frac{1}{x}}}{x}$$
which goes to zero as $x to 0$.
answered Jun 19 '13 at 19:07
Ron GordonRon Gordon
122k14154263
It's easier if you use the plain old addition formula:
$$sin{left(x^2+frac{1}{x}right)} = cos{x^2}, sin{frac{1}{x}} + sin{x^2} ,cos{frac{1}{x}}$$
which behaves as
$$left(1-frac12 x^2right) sin{frac{1}{x}} + x^2 cos{frac{1}{x}}$$
so that
$$frac{sin{left(x^2+frac{1}{x}right)} - sin{frac{1}{x}}}{x} sim frac{x^2 cos{frac{1}{x}} - (1/2)x^2 sin{frac{1}{x}}}{x}$$
which goes to zero as $x to 0$.
answered Jun 19 '13 at 19:07
Ron GordonRon Gordon
122k14154263
answered Jun 19 '13 at 19:07
Ron GordonRon Gordon
122k14154263
answered Jun 19 '13 at 19:07
Ron GordonRon Gordon
122k14154263
answered Jun 19 '13 at 19:07
Ron GordonRon Gordon
122k14154263
122k14154263
add a comment |
add a comment |
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If you have asked this problem before, maybe you could give the link to that previous question?
– celtschk
Jun 19 '13 at 18:18
@user73276:i cant use L'hôpital's rule because $lim_{xto0}{sin(x^2+frac{1}{x})-sinfrac{1}{x}}neq0$
– M.H
Jun 19 '13 at 18:45