How to solve this limit: $\lim\limits_{x \to 0}\left(\frac{(1+2x)^\frac1x}{e^2 +x}\right)^\frac1x$ $$\lim\limits_{x \to 0}\left(\frac{(1+2x)^\frac{1}{x}}{e^2 +x}\right)^\frac{1}{x}=~?$$
Can not solve this limit, already tried with logarithm but this is where i run out of ideas. Thanks.
 A: 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
A: Using L'Hospital rule twice we get $$\lim _{ x\to 0 } \left( \frac { (1+2x)^{ \frac { 1 }{ x }  } }{ e^{ 2 }+x }  \right) ^{ \frac { 1 }{ x }  }=~ { e }^{ \lim _{ x\rightarrow 0 }{ \frac { 1 }{ x } \ln { \left( \frac { (1+2x)^{ \frac { 1 }{ x }  } }{ e^{ 2 }+x }  \right)  }  }  }={ e }^{ \lim _{ x\rightarrow 0 }{ \frac { 1 }{ x } \left[ \frac { 1 }{ x } \ln { \left( 1+2x \right) -\ln { \left( { e }^{ 2 }+x \right)  }  }  \right]  }  }=\\ ={ e }^{ \lim _{ x\rightarrow 0 }{ \frac { \ln { \left( 1+2x \right) -x\ln { \left( { e }^{ 2 }+x \right)  }  }  }{ { x }^{ 2 } }  }  }\overset { L'Hospital }{ = } { e }^{ \lim _{ x\rightarrow 0 }{ \frac { \frac { 2 }{ 1+2x } -\ln { \left( { { e }^{ 2 }+x } \right) -\frac { x }{ { e }^{ 2 }+x }  }  }{ 2{ x } }  }  }\overset { L'hospital }{ = } { e }^{ \lim _{ x\rightarrow 0 }{ \frac { -\frac { 4 }{ { \left( 1+2x \right)  }^{ 2 } } -\frac { 1 }{ { e }^{ 2 }+x } -\frac { { e }^{ 2 } }{ { \left( { e }^{ 2 }+x \right)  }^{ 2 } }  }{ 2 }  }  }={ e }^{ -\frac { 4{ e }^{ 2 }+2 }{ 2{ e }^{ 2 } }  }$$
A: 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.
A: 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:


*

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

*$\ln(1+x) \in x+\lfrac1{2}x^2+O(x^3)$ 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.
A: 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 $\exp(-2-e^{-2})$.
A: 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
$$
\lim_{x\to0}\frac{2x-(2x)^2/2-2x-e^{-2}x^2+o(x^2)}{x^2}=-e^{-2}-2
$$
