Limit of $\frac{1}{x(e^{2/x}-1)}$ as $x \rightarrow \infty$

Question

I am trying to compute the limit of $$\frac{1}{x(e^{2/x}-1)}$$ as $x \rightarrow \infty,$ without using l'Hopital, but I am getting nowhere.

I have tried to rewrite it using $ln()$ so that the expression is $ln\left(\frac{1}{x(e^{2/x}-1))} \right) = ln(1) - ln(x(e^{2/x}-1)) = ln(1) - ln(x) + ln(e^{2/x}-1))$, but that didn't help.

I tried changing a variable but didn't know which one, I tried putting $y=e^{1/x}$ so that $x = \frac{1}{ln(y)}$ and so that the whole expression in the limit is $\frac{ln(y)}{y^2-1}$ as $y \rightarrow 1$ but then the denominator is still $0$.

Any advice?

Edit: This is my first calculus course and I haven't started on taylor expansions, derivatives, integrals or anything like that. So I guess I am supposed to solve it using algebra, basically.

Answer 1

One may use the standard result $$ \lim_{u \to 0}\frac{e^u-1}u=1 $$ giving directly $$ \lim_{x \to \infty}\frac1{x(e^{\large \frac2x}-1)}=\frac12\lim_{x \to \infty}\frac{\large \frac2x}{e^{\large \frac2x}-1}=\frac12 \cdot1=\frac12. $$

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Another path. By the Taylor series expansion, as $u \to 0$, one has $$ e^u=1+u+O(u^2) $$ giving, as $x \to \infty$, $$ e^{\large \frac2x}=1+\frac2x+O\left(\frac1{x^2}\right) $$$$ x(e^{\large \frac2x}-1)=2+O\left(\frac1{x}\right) $$ and, as $x \to \infty$, $$ \frac1{x(e^{\large \frac2x}-1)}=\frac1{2+O\left(\frac1{x}\right)} \to \frac12. $$

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Answer 2

How about $y=1/x$

$$ \lim_{y\rightarrow 0}\frac{y}{e^{2y}-1}$$

Then using the Taylor expansion of $\exp{2y}$

$$ \lim_{y\rightarrow 0}\frac{y}{-1+1+2y+2y^2+\cdots}=\lim_{y\rightarrow 0}\frac{y}{2y}=\frac 12$$

Alternatively, without Taylor: $$ \lim_{y\rightarrow 0}\frac{y}{e^{2y}-1}$$

Take $u=2y$: $$\lim_{y\rightarrow 0}\frac 12 \frac{u}{e^{u}-1}$$

Set $u=\ln(t)$: $$\lim_{t\rightarrow 1}\frac 12 \frac{\ln t}{t-1}$$ Using the definition of $\ln(x)$ $$L=\lim_{t\rightarrow 1}\frac{\ln(t)}{t-1}=\lim_{t\rightarrow 0}\frac{\ln(t+1)}{t}=\lim_{t\rightarrow 0}\int_0^t \frac 1t\frac{dr}{1+r}$$ Put: $r=ts \implies dr=tds$ then: $$L=\lim_{t\rightarrow 0}\int_0^1 \frac{ds}{1+ts}=\int_0^1ds=1$$ P.S.: I know I've used an integral, but can't we apply a simple one like this?