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The limit is this one:

$$\lim_{x \rightarrow 0}\frac{(1+2x)^\frac{1}{x}-(1+x)^\frac{2}{x}}{x}$$

I have found that both $(1+2x)^\frac{1}{x}$ and $(1+x)^\frac{2}{x}$ tend to $e^2$, so the numerator tends to 0. I think that the book said that the result of this limit is $-e^2$ if I recall correctly.

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    can you use L'Hospital? – 2017-02-14
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    @Dr.SonnhardGraubner Yes, I can. I have tried applying L'Hospital, but the derivate of the numerator is pretty complicated and it lead me to nothing helpful – 2017-02-14
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    Limited expansions: for every $(a,b)$, $$(1+ax)^{b/x}=\exp(b\log(1+ax)/x)=\exp(ba-\tfrac12ba^2x+o(x))=e^{ab}(1-\tfrac12ba^2x+o(x))$$ hence $$(1+2x)^{1/x}-(1+x)^{2/x}=e^2(1-\tfrac12\cdot1\cdot2^2x-1+\tfrac12\cdot2\cdot1^2x+o(x))\sim-e^2x$$ hence the limit is indeed $-e^2$. Tools used: when $x\to0$, $$e^x=1+x+o(x)\qquad\log(1+x)=x-\tfrac12x^2+o(x^2)$$ and this is it! – 2017-02-14
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    @Did Sorry, but I don't understand that. I haven't learned about expansions yet, I'm still in high school. – 2017-02-14
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    @user416408 Please list the tools you are allowed to use. – 2017-02-14
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    @Did I only know basic calculus, like derivates, indefinite or definite integrals and limits. That's what I've been taught to date. – 2017-02-14
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    Can't you use Cesaro-Stolz or l'Hosital? – 2017-02-14
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    @theSongbird I can – 2017-02-14
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    What were your results when trying Cesaro-Stolz? – 2017-02-14
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    @theSongbird I don't think I can apply it. From what I know, Stolz-Cesaro is applied when the denominator is strictly monotone and divergent. I have only applied that theorem to limits of sequences, not functions. – 2017-02-14

2 Answers 2

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Note that we use here two well known limits $$\lim _{ \quad x\rightarrow 0 }{ { \left( 1+x \right) }^{ \frac { 1 }{ x } } } =e\\ \lim _{ x\rightarrow 0 }{ \frac { { e }^{ x }-1 }{ x } } =1$$

$$\lim _{ x\rightarrow 0 } \frac { (1+2x)^{ \frac { 1 }{ x } }-(1+x)^{ \frac { 2 }{ x } } }{ x } =\lim _{ x\rightarrow 0 } \frac { { e }^{ \frac { 1 }{ x } \ln { \left( 1+2x \right) } }-{ e }^{ \frac { 2 }{ x } \ln { \left( 1+x \right) } } }{ x } =\\ =\lim _{ x\rightarrow 0 } \frac { { e }^{ \frac { 2 }{ x } \ln { \left( 1+x \right) } }\left[ { e }^{ \frac { 1 }{ x } \ln { \left( 1+2x \right) } -\frac { 2 }{ x } \ln { \left( 1+x \right) } }-1 \right] }{ x } =\\\lim _{ x\rightarrow 0 } \frac { \left[ { e }^{ \frac { 1 }{ x } \ln { \left( 1+2x \right) } -\frac { 2 }{ x } \ln { \left( 1+x \right) } }-1 \right] }{ \frac { 1 }{ x } \ln { \left( 1+2x \right) } -\frac { 2 }{ x } \ln { \left( 1+x \right) } } \cdot \frac { { e }^{ \frac { 2 }{ x } \ln { \left( 1+x \right) } } }{ x } \cdot \left[ \frac { 1 }{ x } \ln { \left( 1+2x \right) } -\frac { 2 }{ x } \ln { \left( 1+x \right) } \right] =\\\ =\lim _{ x\rightarrow 0 } \frac { { e }^{ \frac { 2 }{ x } \ln { \left( 1+x \right) } } }{ x } \cdot \left[ \frac { 1 }{ x } \ln { \left( 1+2x \right) } -\frac { 2 }{ x } \ln { \left( 1+x \right) } \right] =\lim _{ x\rightarrow 0 } \frac { (1+x)^{ \frac { 2 }{ x } } }{ { x }^{ 2 } } \cdot \ln { \left( \frac { 1+2x }{ { \left( 1+x \right) }^{ 2 } } \right) } =\\=\lim _{ x\rightarrow 0 } (1+x)^{ \frac { 2 }{ x } }\cdot \ln { { \left( \frac { 1+2x }{ 1+2x+{ x }^{ 2 } } \right) }^{ \frac { 1 }{ { x }^{ 2 } } } } =\\ =\lim _{ x\rightarrow 0 } (1+x)^{ \frac { 2 }{ x } }\cdot \\\ln { { \left( 1+\frac { -{ x }^{ 2 } }{ 1+2x+{ x }^{ 2 } } \right) }^{ \frac { 1 }{ { x }^{ 2 } } } } =\lim _{ x\rightarrow 0 } (1+x)^{ \frac { 2 }{ x } }\cdot \ln { { \left[ { \left( 1+\frac { -{ x }^{ 2 } }{ 1+2x+{ x }^{ 2 } } \right) }^{ \frac { 1+2x+{ x }^{ 2 } }{ -{ x }^{ 2 } } } \right] }^{ \frac { 1 }{ { x }^{ 2 } } \cdot \frac { -{ x }^{ 2 } }{ 1+2x+{ x }^{ 2 } } } } =\lim _{ x\rightarrow 0 } (1+x)^{ \frac { 2 }{ x } }\cdot \frac { 1 }{ { x }^{ 2 } } \cdot \frac { -{ x }^{ 2 } }{ 1+2x+{ x }^{ 2 } } =\color{red}{-{ e }^{ 2 }}$$

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    Wow! Extraordinary, thank you so much. – 2017-02-14
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    you are welcome) – 2017-02-14
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You can do it with l'Hôpital. Consider $f_{a,b}(x)=(1+ax)^{b/x}$; then $$ \log f_{a,b}(x)=b\frac{\log(1+ax)}{x} $$ and so $$ \frac{f_{a,b}'(x)}{f_{a,b}(x)}=b\frac{\frac{ax}{1+ax}-\log(1+ax)}{x^2}= \frac{b}{1+ax}\frac{ax-(1+ax)\log(1+ax)}{x^2} $$ Since you know that $\lim_{x\to0}f_{a,b}(x)=e^{ab}$, by a basic limit, we have \begin{align} \lim_{x\to0}f_{a,b}'(x) &=be^{ab}\lim_{x\to0}\frac{ax-(1+ax)\log(1+ax)}{x^2}\\ &=be^{ab}\lim_{x\to0}\frac{a-a\log(1+ax)-a}{2x}\\ &=-\frac{ab}{2}e^{ab}\lim_{x\to0}\frac{\log(1+ax)}{x}\\ &=-\frac{a^2b}{2}e^{ab} \end{align} Thus we have $$ \lim_{x\to0}\frac{f_{2,1}(x)-f_{1,2}(x)}{x}= \lim_{x\to0}\bigl(f_{2,1}'(x)-f_{1,2}'(x)\bigr)= -2e^2+e^2=-e^2 $$

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    Thanks for answering. I'm sorry, but I don't know what you mean by those a and b indexes. – 2017-02-14
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    @user416408 The function $(1+2x)^{1/x}$ is $f_{a,b}$ for $a=2$ and $b=1$. It's just a family of functions. Ease of computation: no need to do twice the same computations, as the other part is $f_{a,b}$ for $a=1$ and $b=2$. – 2017-02-14
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    Thank you. I understand now. – 2017-02-15