How to calculate $ lim_{xto 0} (frac{(5x + 1)^{20} - (20x + 1)^{5}}{sqrt[5]{1 + 20x^{2}}-1}) $ without the...
$begingroup$
$$
lim_{xto 0} (frac{(5x + 1)^{20} - (20x + 1)^{5}}{sqrt[5]{1 + 20x^{2}}-1})
$$
Hello! I need to solve this limit. I had solved it with the rule of L'Hôpital, but i can't without it. I tried multiplicatio using Special Limits, but i simplified it only to $$ lim_{xto 0} (frac{e^{100x} - e^{100x}}{e^{4x^2}-1})$$ So i think i had done something wrong and i should do it by another way. Please help me, I must solve it using only Special Limits and simple transformations. I can't use derivatives.
calculus sequences-and-series limits limits-without-lhopital
asked Dec 24 '18 at 18:38
Влад СивиринВлад Сивирин
1086
$endgroup$
add a comment |
$begingroup$
$$
lim_{xto 0} (frac{(5x + 1)^{20} - (20x + 1)^{5}}{sqrt[5]{1 + 20x^{2}}-1})
$$
Hello! I need to solve this limit. I had solved it with the rule of L'Hôpital, but i can't without it. I tried multiplicatio using Special Limits, but i simplified it only to $$ lim_{xto 0} (frac{e^{100x} - e^{100x}}{e^{4x^2}-1})$$ So i think i had done something wrong and i should do it by another way. Please help me, I must solve it using only Special Limits and simple transformations. I can't use derivatives.
calculus sequences-and-series limits limits-without-lhopital
asked Dec 24 '18 at 18:38
Влад СивиринВлад Сивирин
1086
$endgroup$
1
$begingroup$
Why on Earth would you want to evaluate THAT?
$endgroup$
– Zachary Selk
Dec 24 '18 at 18:45
1
$begingroup$
You couldn't have simplified it to that because "that" is $0.$
$endgroup$
– zhw.
Dec 24 '18 at 18:53
$begingroup$
For the numerator use binomial theorem to get $(190cdot 25-10cdot 400)x^2+o(x^2)$ and for denominator use the standard limit $limlimits_{xto a} dfrac{x^n-a^n} {x-a} =na^{n-1}$.
$endgroup$
– Paramanand Singh
Dec 25 '18 at 3:48
add a comment |
$begingroup$
$$
lim_{xto 0} (frac{(5x + 1)^{20} - (20x + 1)^{5}}{sqrt[5]{1 + 20x^{2}}-1})
$$
Hello! I need to solve this limit. I had solved it with the rule of L'Hôpital, but i can't without it. I tried multiplicatio using Special Limits, but i simplified it only to $$ lim_{xto 0} (frac{e^{100x} - e^{100x}}{e^{4x^2}-1})$$ So i think i had done something wrong and i should do it by another way. Please help me, I must solve it using only Special Limits and simple transformations. I can't use derivatives.
calculus sequences-and-series limits limits-without-lhopital
asked Dec 24 '18 at 18:38
Влад СивиринВлад Сивирин
1086
$endgroup$
$$
lim_{xto 0} (frac{(5x + 1)^{20} - (20x + 1)^{5}}{sqrt[5]{1 + 20x^{2}}-1})
$$
Hello! I need to solve this limit. I had solved it with the rule of L'Hôpital, but i can't without it. I tried multiplicatio using Special Limits, but i simplified it only to $$ lim_{xto 0} (frac{e^{100x} - e^{100x}}{e^{4x^2}-1})$$ So i think i had done something wrong and i should do it by another way. Please help me, I must solve it using only Special Limits and simple transformations. I can't use derivatives.
calculus sequences-and-series limits limits-without-lhopital
calculus sequences-and-series limits limits-without-lhopital
asked Dec 24 '18 at 18:38
Влад СивиринВлад Сивирин
1086
asked Dec 24 '18 at 18:38
Влад СивиринВлад Сивирин
1086
edited Dec 24 '18 at 19:06
Влад Сивирин
asked Dec 24 '18 at 18:38
Влад СивиринВлад Сивирин
1086
asked Dec 24 '18 at 18:38
Влад СивиринВлад Сивирин
1086
asked Dec 24 '18 at 18:38
Влад СивиринВлад Сивирин
1086
1086
1
$begingroup$
Why on Earth would you want to evaluate THAT?
$endgroup$
– Zachary Selk
Dec 24 '18 at 18:45
1
$begingroup$
You couldn't have simplified it to that because "that" is $0.$
$endgroup$
– zhw.
Dec 24 '18 at 18:53
$begingroup$
For the numerator use binomial theorem to get $(190cdot 25-10cdot 400)x^2+o(x^2)$ and for denominator use the standard limit $limlimits_{xto a} dfrac{x^n-a^n} {x-a} =na^{n-1}$.
$endgroup$
– Paramanand Singh
Dec 25 '18 at 3:48
add a comment |
1
$begingroup$
Why on Earth would you want to evaluate THAT?
$endgroup$
– Zachary Selk
Dec 24 '18 at 18:45
1
$begingroup$
You couldn't have simplified it to that because "that" is $0.$
$endgroup$
– zhw.
Dec 24 '18 at 18:53
$begingroup$
For the numerator use binomial theorem to get $(190cdot 25-10cdot 400)x^2+o(x^2)$ and for denominator use the standard limit $limlimits_{xto a} dfrac{x^n-a^n} {x-a} =na^{n-1}$.
$endgroup$
– Paramanand Singh
Dec 25 '18 at 3:48
1
1
$begingroup$
Why on Earth would you want to evaluate THAT?
$endgroup$
– Zachary Selk
Dec 24 '18 at 18:45
$begingroup$
Why on Earth would you want to evaluate THAT?
$endgroup$
– Zachary Selk
Dec 24 '18 at 18:45
1
1
$begingroup$
You couldn't have simplified it to that because "that" is $0.$
$endgroup$
– zhw.
Dec 24 '18 at 18:53
$begingroup$
You couldn't have simplified it to that because "that" is $0.$
$endgroup$
– zhw.
Dec 24 '18 at 18:53
$begingroup$
For the numerator use binomial theorem to get $(190cdot 25-10cdot 400)x^2+o(x^2)$ and for denominator use the standard limit $limlimits_{xto a} dfrac{x^n-a^n} {x-a} =na^{n-1}$.
$endgroup$
– Paramanand Singh
Dec 25 '18 at 3:48
$begingroup$
For the numerator use binomial theorem to get $(190cdot 25-10cdot 400)x^2+o(x^2)$ and for denominator use the standard limit $limlimits_{xto a} dfrac{x^n-a^n} {x-a} =na^{n-1}$.
$endgroup$
– Paramanand Singh
Dec 25 '18 at 3:48
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You can just use the Taylor series on the bottom and expand the top, keeping only the first terms that do not cancel.
$$(5x + 1)^{20} - (20x + 1)^{5}=1+100x+20cdot 19 cdot frac 12cdot 5^2 x^2+O(x^3)-1-100x-5cdot 4 cdot frac 12cdot 20^2+O(x^3)\=750x^2+O(x^3)\
sqrt[5]{1 + 20x^{2}}-1=1+frac {20x^2}5+O(x^4)-1=4x^2+O(x^4)$$
and the ratio is just $$frac {750}4+O(x)$$
answered Dec 24 '18 at 18:50
Ross MillikanRoss Millikan
295k23198371
$endgroup$
add a comment |
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1 Answer
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$begingroup$
You can just use the Taylor series on the bottom and expand the top, keeping only the first terms that do not cancel.
$$(5x + 1)^{20} - (20x + 1)^{5}=1+100x+20cdot 19 cdot frac 12cdot 5^2 x^2+O(x^3)-1-100x-5cdot 4 cdot frac 12cdot 20^2+O(x^3)\=750x^2+O(x^3)\
sqrt[5]{1 + 20x^{2}}-1=1+frac {20x^2}5+O(x^4)-1=4x^2+O(x^4)$$
and the ratio is just $$frac {750}4+O(x)$$
answered Dec 24 '18 at 18:50
Ross MillikanRoss Millikan
295k23198371
$endgroup$
add a comment |
$begingroup$
You can just use the Taylor series on the bottom and expand the top, keeping only the first terms that do not cancel.
$$(5x + 1)^{20} - (20x + 1)^{5}=1+100x+20cdot 19 cdot frac 12cdot 5^2 x^2+O(x^3)-1-100x-5cdot 4 cdot frac 12cdot 20^2+O(x^3)\=750x^2+O(x^3)\
sqrt[5]{1 + 20x^{2}}-1=1+frac {20x^2}5+O(x^4)-1=4x^2+O(x^4)$$
and the ratio is just $$frac {750}4+O(x)$$
answered Dec 24 '18 at 18:50
Ross MillikanRoss Millikan
295k23198371
$endgroup$
add a comment |
$begingroup$
You can just use the Taylor series on the bottom and expand the top, keeping only the first terms that do not cancel.
$$(5x + 1)^{20} - (20x + 1)^{5}=1+100x+20cdot 19 cdot frac 12cdot 5^2 x^2+O(x^3)-1-100x-5cdot 4 cdot frac 12cdot 20^2+O(x^3)\=750x^2+O(x^3)\
sqrt[5]{1 + 20x^{2}}-1=1+frac {20x^2}5+O(x^4)-1=4x^2+O(x^4)$$
and the ratio is just $$frac {750}4+O(x)$$
answered Dec 24 '18 at 18:50
Ross MillikanRoss Millikan
295k23198371
$endgroup$
You can just use the Taylor series on the bottom and expand the top, keeping only the first terms that do not cancel.
$$(5x + 1)^{20} - (20x + 1)^{5}=1+100x+20cdot 19 cdot frac 12cdot 5^2 x^2+O(x^3)-1-100x-5cdot 4 cdot frac 12cdot 20^2+O(x^3)\=750x^2+O(x^3)\
sqrt[5]{1 + 20x^{2}}-1=1+frac {20x^2}5+O(x^4)-1=4x^2+O(x^4)$$
and the ratio is just $$frac {750}4+O(x)$$
answered Dec 24 '18 at 18:50
Ross MillikanRoss Millikan
295k23198371
answered Dec 24 '18 at 18:50
Ross MillikanRoss Millikan
295k23198371
answered Dec 24 '18 at 18:50
Ross MillikanRoss Millikan
295k23198371
answered Dec 24 '18 at 18:50
Ross MillikanRoss Millikan
295k23198371
295k23198371
add a comment |
add a comment |
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1
$begingroup$
Why on Earth would you want to evaluate THAT?
$endgroup$
– Zachary Selk
Dec 24 '18 at 18:45
1
$begingroup$
You couldn't have simplified it to that because "that" is $0.$
$endgroup$
– zhw.
Dec 24 '18 at 18:53
$begingroup$
For the numerator use binomial theorem to get $(190cdot 25-10cdot 400)x^2+o(x^2)$ and for denominator use the standard limit $limlimits_{xto a} dfrac{x^n-a^n} {x-a} =na^{n-1}$.
$endgroup$
– Paramanand Singh
Dec 25 '18 at 3:48