# How to solve this limit: $\lim\limits_{x \to 0}\left(\frac{(1+2x)^\frac1x}{e^2 +x}\right)^\frac1x$ [closed]

$$\lim\limits_{x \to 0}\left(\frac{(1+2x)^\frac{1}{x}}{e^2 +x}\right)^\frac{1}{x}=~?$$

Can not solve this limit, already tried with logarithm but this is where i run out of ideas. Thanks.

• Convert $(1+2x)^\frac{1}{x}$ to $e$. – Nosrati Jan 29 '17 at 12:14
• @MyGlasses To $e^2$. – Mycroft Jan 29 '17 at 12:16
• @AlanTuring Of course. – Nosrati Jan 29 '17 at 12:19

HINT: write $$y=\left(\frac{(1+2x)^{1/x}}{e^2+x}\right)^{1/x}$$ and take the logarithm on both sides and write $$e^{\frac{\ln\left(\frac{(1+2x)^{1/x}}{e^2+x}\right)}{x}}$$ and use the rules of L'Hospital

Using L'Hospital rule twice we get $$\lim _{ x\to 0 } \left( \frac { (1+2x)^{ \frac { 1 }{ x } } }{ e^{ 2 }+x } \right) ^{ \frac { 1 }{ x } }=~ { e }^{ \lim _{ x\rightarrow 0 }{ \frac { 1 }{ x } \ln { \left( \frac { (1+2x)^{ \frac { 1 }{ x } } }{ e^{ 2 }+x } \right) } } }={ e }^{ \lim _{ x\rightarrow 0 }{ \frac { 1 }{ x } \left[ \frac { 1 }{ x } \ln { \left( 1+2x \right) -\ln { \left( { e }^{ 2 }+x \right) } } \right] } }=\\ ={ e }^{ \lim _{ x\rightarrow 0 }{ \frac { \ln { \left( 1+2x \right) -x\ln { \left( { e }^{ 2 }+x \right) } } }{ { x }^{ 2 } } } }\overset { L'Hospital }{ = } { e }^{ \lim _{ x\rightarrow 0 }{ \frac { \frac { 2 }{ 1+2x } -\ln { \left( { { e }^{ 2 }+x } \right) -\frac { x }{ { e }^{ 2 }+x } } }{ 2{ x } } } }\overset { L'hospital }{ = } { e }^{ \lim _{ x\rightarrow 0 }{ \frac { -\frac { 4 }{ { \left( 1+2x \right) }^{ 2 } } -\frac { 1 }{ { e }^{ 2 }+x } -\frac { { e }^{ 2 } }{ { \left( { e }^{ 2 }+x \right) }^{ 2 } } }{ 2 } } }={ e }^{ -\frac { 4{ e }^{ 2 }+2 }{ 2{ e }^{ 2 } } }$$

Let us first examine the inner part.

$$\lim_{x \to 0} \frac{(1+2x)^{\frac{1}{x}}}{e^2+x}$$

It is easy to see that numerator is of the form $1^{ \infty}$ So the numerator goes to $e^2$.

Rewritting the limit $$\lim_{x \to 0}\frac{1}{e^{2/x}}(\frac{e^2}{e^2+x})^{\frac{1}{x}} (1+2x)^{1/x^2}$$ which can be rewritten as $$e^{-e^{-2}} \lim_{x \to 0}[\frac{(1+2x)^{1/x}}{e^2}]^{1/x}$$

Now evaluating the limit under the inner part.take ln both sides and use Lhospital rule we get the limit as $$e^{-2-e^{-2}}$$.In other way notice that it is in the form $1^{\infty}$.So the limit can be written as $$e^{\lim_{x \to 0} \frac{(1+2x)^{1/x}-e^2}{e^2}(1/x)}$$.Now application of lhospitals rule becomes easier.

All such limits can be mechanically computed easily using asymptotic expansions. One should not use L'Hopital unless it is obvious that it works well. $\def\lfrac#1#2{{\large\frac{#1}{#2}}} \def\wi{\subseteq}$

The complete solution produced by the mechanical computation is as follows.

As $x \to 0$:

$\Big(\lfrac{(1+2x)^{1/x}}{e^2+x}\Big)^{1/x}$ $= \Big(\lfrac{\exp(\lfrac1x\ln(1+2x))}{e^2+x}\Big)^{1/x}$ $\in \Big(\lfrac{\exp(\lfrac1x(2x-2x^2+O(x^3)))}{e^2+x}\Big)^{1/x}$

$= \Big(\lfrac{\exp(2-2x+O(x^2))}{e^2+x}\Big)^{1/x}$ $= e^{-2} \Big(\lfrac{\exp(O(x^2))}{1+e^{-2}x}\Big)^{1/x}$ $\wi e^{-2} \Big(\lfrac{1+O(x^2)}{1+e^{-2}x}\Big)^{1/x}$

$\wi e^{-2} \Big((1+O(x^2))(1-e^{-2}x)\Big)^{1/x}$ $\wi e^{-2} \Big(1-e^{-2}x+O(x^2)\Big)^{1/x}$

$= e^{-2} \exp\!\Big(\lfrac1x\ln(1-e^{-2}x+O(x^2))\Big)$ $\wi e^{-2} \exp\!\Big(\lfrac1x(-e^{-2}x+O(x^2))\Big)$

$= e^{-2} \exp(-e^{-2}+O(x))$ $= e^{-2-e^{-2}} \exp(O(x))$ $\wi e^{-2-e^{-2}}(1+O(x))$

$= e^{-2-e^{-2}}+O(x)$.

The two asymptotic expansions used in the above solution are:

1. $\exp(x) \in 1+O(x)$ if $x \in o(1)$.

2. $\ln(1+x) \in x+\lfrac1{2}x^2+O(x^3)$ if $x \in o(1)$.

One question is how I know how many terms of the asymptotic expansions to use. The answer is I do not know in advance. I just start with using the first two terms, and if at some point I cannot simplify due to the terms cancelling and leaving only asymptotic classes, then I trace where those remaining terms arose from and increase the number of terms in previous asymptotic expansions where needed. This is a mechanical process and is actually used in computer algebra systems to find such limits.

Let the expression be denoted by $u^{v}$ where $u\to 1$ as $x\to 0$. If $L$ is the desired limit then $\log L = \lim_{x\to 0}v\log u$ and then using the fact that $(\log u) /(u - 1)\to 1$ we see that $\log L =\lim_{x\to 0}v(u-1)$. Next note that $u$ is of the form $a/b$ where both $a, b$ tend to $e^{2}$ and hence $\log L=e^{-2}\lim_{x\to 0}v(a-b)$. We can see that $$v(a - b) = \frac{(1+2x)^{1/x} - e^{2}}{x} - 1= e^{2}\cdot \frac{\exp(t)-1}{x} - 1$$ where $$t = \frac{\log(1+2x)}{x} - 2\to 0$$ and then using the fact that $(\exp(t) - 1)/t\to 1$ we see that $$\lim_{x\to 0}v(a-b)= e^{2}\lim_{x\to 0}\frac{t}{x} - 1$$ Using Taylor/L'Hospital's Rule we can see that $$\frac{t} {x} =\frac{\log(1+2x)-2x}{x^{2}}\to - 2$$ and hence $v(a-b) \to - 2e^{2}-1$ and hence $\log L=-2-e^{-2}$. Therefore the desired limit is $\exp(-2-e^{-2})$.

The idea of taking the logarithm is good: $$\lim_{x\to0}\frac{\dfrac{\log(1+2x)}{x}-\log(e^2+x)}{x}= \lim_{x\to0}\frac{\log(1+2x)-x\log(e^2+x)}{x^2}$$ Now let's compute the Taylor expansion of $\log(e^2+x)$: $$\log(e^2+x)=2+\log(1+e^{-2}x)=2+\frac{x}{e^2}+o(x)$$ so we have $$\lim_{x\to0}\frac{2x-(2x)^2/2-2x-e^{-2}x^2+o(x^2)}{x^2}=-e^{-2}-2$$