limit as x goes to zero

Question

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

Can anyone help with this limit? I know I have to split it up and apply L'hopitals rule, and the answer is $$\frac{-2}{e}$$ I just cant get it to work out on paper.

Answer 1

hint

The numerator can be set as

$$x( x\sin(\frac 1x) +1 )$$

observe that $$\lim_{x\to 0}\Bigl(x\sin(\frac 1x) +1 \Bigr) =1$$

now, we need to compute $$\lim_{x\to 0}\frac{x}{(1+x)^\frac 1x -e}$$ with $$(1+x)^\frac 1x=e^{\frac 1x\ln(1+x)}=e^{f(x)}.$$

now use l'Hospital to get

$$\lim_{x\to 0}\frac{1}{f'(x)e^{f(x)}}$$

where $$f'(x)=\frac{\frac{x}{x+1}-\ln(1+x)}{x^2}$$

Answer 2

Interesting limit. After the simplification already proposed you get to $$\mathcal L =\lim_{x\to 0}\frac{x}{(1+x)^{\frac1{x}}-e},$$ which, by the way, could be interpreted as the inverse of the derivative in $0$ of the function $$f(x)=\begin{cases}(1+x)^{\frac1{x}} & (x>-1 \land x\neq 0)\\e& (x=0).\end{cases}$$

I would go like that \begin{eqnarray} \mathcal L &=& \lim_{x\to 0}\frac{x}{e^{\frac1{x}\log(1+x)}-e}=\\ &=&\lim_{x\to 0}\frac{x}{e\left[e^{\frac1{x}\log(1+x)-1}-1\right]}. \end{eqnarray} Now use $$\frac{\log(1+\alpha(x))}{\alpha(x)} \to 1$$ and $$\frac{e^{\alpha(x)}-1}{\alpha(x)}\to 1$$ when $\alpha(x) \to 0$, to get \begin{eqnarray} \mathcal L &=& \frac1e\cdot\lim_{x\to 0}\frac{x}{\frac1{x}\log(1+x)-1}=\\ &=&\frac1e\cdot\lim_{x\to 0} \frac{x^2}{\log(1+x)-x}. \end{eqnarray} Use now MacLaurin or de l'Hospital to handle it.

Answer 3

The exercise is meant to make you not use l'Hôpital, because the derivative of the numerator has no limit at $0$. It's not difficult to do it with Taylor expansion.

You know that $$ \frac{1}{x}\log(1+x)=\frac{1}{x}(x-x^2/2+o(x^2))=1-x/2+o(x) $$ so $$ (1+x)^{1/x}-e=e^{1-x/2+o(x)}-e=e\cdot e^{-x/2+o(x)}-e=e(1-x/2+o(x))-e=-ex/2+o(x) $$ Note also that $$ x^2\sin\frac{1}{x}=o(x) $$ and therefore the limit is $$ \lim_{x\to0}\frac{x+o(x)}{-ex/2+o(x)}=-\frac{2}{e} $$