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Solving without expansions in limits

Help with solving limitsA problem in limitsFinding the limit without L'Hospital's rule.Solving Limits with L'Hospital's RulesEvaluate limits of trigonometric equationHow to calculate $limlimits_{xto 0} frac{[sin{x}-x][cos({3x})-1]}{x[e^x -1]^4}$ without using L'Hôpital's Rule?Evaluating limits via Infinite SeriesExamples of limits that become easier with Taylor seriesEvaluating $lim limits_{x to 0} frac {e^{-1/x^2}}{x}$ without using L' Hôpital's ruleClarify difference in solving limits to infinity vs finite limts

3

3

$begingroup$

Evaluate $limlimits_{x to 0} frac{(1+x)^{1/x} - e + frac{1}{2}ex}{x^2}$

One way that I can immediately think of is expanding each of the terms and solving like,
$$(1+x)^{1/x} = e^{log_e (1+x)^{1/x}} = e^{frac{1}{x} (x-frac{x^2}{2} -frac{x^3}{3}+...)}$$
and then after complete expansion of each and every and substuting into back to limit and solving I get $frac{11e}{24}$ as an answer.

Now, this is a relatively long and complicated way to solve as you can see. I want to know if there is an easier way to solve this problem. Please help. Thank you!

calculus limits

share|cite|improve this question

edited 4 hours ago

rash

asked 4 hours ago

rashrash

34812

$endgroup$

add a comment | 

3

3

$begingroup$

Evaluate $limlimits_{x to 0} frac{(1+x)^{1/x} - e + frac{1}{2}ex}{x^2}$

One way that I can immediately think of is expanding each of the terms and solving like,
$$(1+x)^{1/x} = e^{log_e (1+x)^{1/x}} = e^{frac{1}{x} (x-frac{x^2}{2} -frac{x^3}{3}+...)}$$
and then after complete expansion of each and every and substuting into back to limit and solving I get $frac{11e}{24}$ as an answer.

Now, this is a relatively long and complicated way to solve as you can see. I want to know if there is an easier way to solve this problem. Please help. Thank you!

calculus limits

share|cite|improve this question

edited 4 hours ago

rash

asked 4 hours ago

rashrash

34812

$endgroup$

add a comment | 

$begingroup$

Evaluate $limlimits_{x to 0} frac{(1+x)^{1/x} - e + frac{1}{2}ex}{x^2}$

One way that I can immediately think of is expanding each of the terms and solving like,
$$(1+x)^{1/x} = e^{log_e (1+x)^{1/x}} = e^{frac{1}{x} (x-frac{x^2}{2} -frac{x^3}{3}+...)}$$
and then after complete expansion of each and every and substuting into back to limit and solving I get $frac{11e}{24}$ as an answer.

Now, this is a relatively long and complicated way to solve as you can see. I want to know if there is an easier way to solve this problem. Please help. Thank you!

calculus limits

share|cite|improve this question

edited 4 hours ago

rash

asked 4 hours ago

rashrash

34812

$endgroup$

share|cite|improve this question

edited 4 hours ago

rash

asked 4 hours ago

rashrash

34812

share|cite|improve this question

edited 4 hours ago

rash

asked 4 hours ago

rashrash

34812

asked 4 hours ago

rashrash

34812

asked 4 hours ago

rashrash

34812

2 Answers
2

active

oldest

votes

1

$begingroup$

Write $f(x) = frac{1}{x}log(1+x)$ and $f(0) = 1$. We know that $f$ so defined is analytics near $0$. Now, by the L'Hospital's rule applied twice,

begin{align*}
lim_{xto0} frac{e^{f(x)} - e + frac{e}{2}x}{x^2}
&= lim_{xto0} frac{e^{f(x)}f'(x) + frac{e}{2}}{2x} \
&= lim_{xto0} frac{e^{f(x)}f''(x) + e^{f(x)}f'(x)^2}{2} \
&= frac{e}{2}f''(0) + frac{e}{2}f'(0)^2.
end{align*}

Since $ f(x) = 1 - frac{1}{2}x + frac{1}{3}x^2 + cdots $ near $0$, it follows that $f'(0) = -frac{1}{2}$ and $f''(0) = frac{2}{3}$. Therefore the limit equals

$$ frac{e}{2}cdotfrac{2}{3} + frac{e}{2}left(-frac{1}{2}right)^2
= frac{11}{24}e. $$

share|cite|improve this answer

answered 19 mins ago

Sangchul LeeSangchul Lee

95.6k12171279

$endgroup$

add a comment | 

2

$begingroup$

If I may suggest, the problem of
$$y=frac{(1+x)^{frac1 x} - e + frac{1}{2}ex}{x^2}$$ is not so difficult if you use another way.
$$a=(1+x)^{frac1 x}implies log(a)= {frac1 x}log(1+x)$$
$$ log(a)={frac1 x}left(x-frac{x^2}{2}+frac{x^3}{3}-frac{x^4}{4}+Oleft(x^5right) right)=1-frac{x}{2}+frac{x^2}{3}-frac{x^3}{4}+Oleft(x^4right)$$ Now, continuing with Taylor
$$a=e^{log(a)}=e-frac{e x}{2}+frac{11 e x^2}{24}-frac{7 e x^3}{16}+Oleft(x^4right)$$
$$y=frac{frac{11 e x^2}{24}-frac{7 e x^3}{16}+Oleft(x^4right) }{x^2}=frac{11 e}{24}-frac{7 e x}{16}+Oleft(x^2right)$$ which gives not only the limit but also how it is approached.

share|cite|improve this answer

answered 2 hours ago

Claude LeiboviciClaude Leibovici

124k1157135

$endgroup$

add a comment | 

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1

$begingroup$

Write $f(x) = frac{1}{x}log(1+x)$ and $f(0) = 1$. We know that $f$ so defined is analytics near $0$. Now, by the L'Hospital's rule applied twice,

begin{align*}
lim_{xto0} frac{e^{f(x)} - e + frac{e}{2}x}{x^2}
&= lim_{xto0} frac{e^{f(x)}f'(x) + frac{e}{2}}{2x} \
&= lim_{xto0} frac{e^{f(x)}f''(x) + e^{f(x)}f'(x)^2}{2} \
&= frac{e}{2}f''(0) + frac{e}{2}f'(0)^2.
end{align*}

Since $ f(x) = 1 - frac{1}{2}x + frac{1}{3}x^2 + cdots $ near $0$, it follows that $f'(0) = -frac{1}{2}$ and $f''(0) = frac{2}{3}$. Therefore the limit equals

$$ frac{e}{2}cdotfrac{2}{3} + frac{e}{2}left(-frac{1}{2}right)^2
= frac{11}{24}e. $$

share|cite|improve this answer

answered 19 mins ago

Sangchul LeeSangchul Lee

95.6k12171279

$endgroup$

add a comment | 

1

$begingroup$

Write $f(x) = frac{1}{x}log(1+x)$ and $f(0) = 1$. We know that $f$ so defined is analytics near $0$. Now, by the L'Hospital's rule applied twice,

begin{align*}
lim_{xto0} frac{e^{f(x)} - e + frac{e}{2}x}{x^2}
&= lim_{xto0} frac{e^{f(x)}f'(x) + frac{e}{2}}{2x} \
&= lim_{xto0} frac{e^{f(x)}f''(x) + e^{f(x)}f'(x)^2}{2} \
&= frac{e}{2}f''(0) + frac{e}{2}f'(0)^2.
end{align*}

Since $ f(x) = 1 - frac{1}{2}x + frac{1}{3}x^2 + cdots $ near $0$, it follows that $f'(0) = -frac{1}{2}$ and $f''(0) = frac{2}{3}$. Therefore the limit equals

$$ frac{e}{2}cdotfrac{2}{3} + frac{e}{2}left(-frac{1}{2}right)^2
= frac{11}{24}e. $$

share|cite|improve this answer

answered 19 mins ago

Sangchul LeeSangchul Lee

95.6k12171279

$endgroup$

add a comment | 

$begingroup$

Write $f(x) = frac{1}{x}log(1+x)$ and $f(0) = 1$. We know that $f$ so defined is analytics near $0$. Now, by the L'Hospital's rule applied twice,

begin{align*}
lim_{xto0} frac{e^{f(x)} - e + frac{e}{2}x}{x^2}
&= lim_{xto0} frac{e^{f(x)}f'(x) + frac{e}{2}}{2x} \
&= lim_{xto0} frac{e^{f(x)}f''(x) + e^{f(x)}f'(x)^2}{2} \
&= frac{e}{2}f''(0) + frac{e}{2}f'(0)^2.
end{align*}

Since $ f(x) = 1 - frac{1}{2}x + frac{1}{3}x^2 + cdots $ near $0$, it follows that $f'(0) = -frac{1}{2}$ and $f''(0) = frac{2}{3}$. Therefore the limit equals

$$ frac{e}{2}cdotfrac{2}{3} + frac{e}{2}left(-frac{1}{2}right)^2
= frac{11}{24}e. $$

share|cite|improve this answer

answered 19 mins ago

Sangchul LeeSangchul Lee

95.6k12171279

$endgroup$

share|cite|improve this answer

answered 19 mins ago

Sangchul LeeSangchul Lee

95.6k12171279

answered 19 mins ago

Sangchul LeeSangchul Lee

95.6k12171279

answered 19 mins ago

Sangchul LeeSangchul Lee

95.6k12171279

2

$begingroup$

If I may suggest, the problem of
$$y=frac{(1+x)^{frac1 x} - e + frac{1}{2}ex}{x^2}$$ is not so difficult if you use another way.
$$a=(1+x)^{frac1 x}implies log(a)= {frac1 x}log(1+x)$$
$$ log(a)={frac1 x}left(x-frac{x^2}{2}+frac{x^3}{3}-frac{x^4}{4}+Oleft(x^5right) right)=1-frac{x}{2}+frac{x^2}{3}-frac{x^3}{4}+Oleft(x^4right)$$ Now, continuing with Taylor
$$a=e^{log(a)}=e-frac{e x}{2}+frac{11 e x^2}{24}-frac{7 e x^3}{16}+Oleft(x^4right)$$
$$y=frac{frac{11 e x^2}{24}-frac{7 e x^3}{16}+Oleft(x^4right) }{x^2}=frac{11 e}{24}-frac{7 e x}{16}+Oleft(x^2right)$$ which gives not only the limit but also how it is approached.

share|cite|improve this answer

answered 2 hours ago

Claude LeiboviciClaude Leibovici

124k1157135

$endgroup$

add a comment | 

2

$begingroup$

If I may suggest, the problem of
$$y=frac{(1+x)^{frac1 x} - e + frac{1}{2}ex}{x^2}$$ is not so difficult if you use another way.
$$a=(1+x)^{frac1 x}implies log(a)= {frac1 x}log(1+x)$$
$$ log(a)={frac1 x}left(x-frac{x^2}{2}+frac{x^3}{3}-frac{x^4}{4}+Oleft(x^5right) right)=1-frac{x}{2}+frac{x^2}{3}-frac{x^3}{4}+Oleft(x^4right)$$ Now, continuing with Taylor
$$a=e^{log(a)}=e-frac{e x}{2}+frac{11 e x^2}{24}-frac{7 e x^3}{16}+Oleft(x^4right)$$
$$y=frac{frac{11 e x^2}{24}-frac{7 e x^3}{16}+Oleft(x^4right) }{x^2}=frac{11 e}{24}-frac{7 e x}{16}+Oleft(x^2right)$$ which gives not only the limit but also how it is approached.

share|cite|improve this answer

answered 2 hours ago

Claude LeiboviciClaude Leibovici

124k1157135

$endgroup$

add a comment | 

$begingroup$

If I may suggest, the problem of
$$y=frac{(1+x)^{frac1 x} - e + frac{1}{2}ex}{x^2}$$ is not so difficult if you use another way.
$$a=(1+x)^{frac1 x}implies log(a)= {frac1 x}log(1+x)$$
$$ log(a)={frac1 x}left(x-frac{x^2}{2}+frac{x^3}{3}-frac{x^4}{4}+Oleft(x^5right) right)=1-frac{x}{2}+frac{x^2}{3}-frac{x^3}{4}+Oleft(x^4right)$$ Now, continuing with Taylor
$$a=e^{log(a)}=e-frac{e x}{2}+frac{11 e x^2}{24}-frac{7 e x^3}{16}+Oleft(x^4right)$$
$$y=frac{frac{11 e x^2}{24}-frac{7 e x^3}{16}+Oleft(x^4right) }{x^2}=frac{11 e}{24}-frac{7 e x}{16}+Oleft(x^2right)$$ which gives not only the limit but also how it is approached.

share|cite|improve this answer

answered 2 hours ago

Claude LeiboviciClaude Leibovici

124k1157135

$endgroup$

share|cite|improve this answer

answered 2 hours ago

Claude LeiboviciClaude Leibovici

124k1157135

answered 2 hours ago

Claude LeiboviciClaude Leibovici

124k1157135

answered 2 hours ago

Claude LeiboviciClaude Leibovici

124k1157135