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

I tried factoring $x$ then writing it as $1/x$ and applying l'Hospital, I got the answer $3$ but my book say it should be $1$.

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  • $\begingroup$ i fixed it , this is the limit $\endgroup$ – oren revenge Mar 26 '16 at 11:17
  • $\begingroup$ it should be one. $\endgroup$ – Dhanush Krishna Mar 26 '16 at 11:19
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What are you allowed to do? The easiest is to expand $e^{-\frac{1}{x}} \sim 1-\frac{1}{x} + \frac{1}{2x^2}$

EDIT: Ok, another way is to rewrite it and set $\frac{1}{x} = t$: $$ \lim_{t \to 0} \frac{e^{-t}-1}{t} + 2e^{-t} $$ and recognize the definition of a derivative in the first term and take the limit

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  • $\begingroup$ How do I know to expand in that form? $\endgroup$ – oren revenge Mar 26 '16 at 11:23
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    $\begingroup$ This is Maclaurin series expansion, you expand $e^t$ as $t$ gets close to 0 $\endgroup$ – Alex Mar 26 '16 at 11:26
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Like Ennar, set $1/x=t$ to get $$\lim_{t\to0^+}\dfrac{(1+2t)-e^t}t\cdot\dfrac1{\lim_{t\to0^+} e^t} =2-\lim_{t\to0^+}\dfrac{e^t-1}t=?$$

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  • $\begingroup$ Very nice, the last limit is $\left.\frac d{dt}(e^t)\right|_{t=0} = 1$, +1 for no l'Hospital. $\endgroup$ – Ennar Mar 26 '16 at 14:45
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As $x\to\infty$, $$ e^{-1/x}=1-\frac1x+O\left(\frac1{x^2}\right) $$ So $$ \begin{align} (x+2)e^{-1/x}-x &=x+1+O\left(\frac1x\right)-x\\ &=1+O\left(\frac1x\right) \end{align} $$

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If you let $t = \frac 1 x$, then \begin{align} \lim_{x\to\infty}(x+2)e^{-\frac 1 x}-x &= \lim_{t\to 0^+}(\frac 1 t + 2)e^{-t}-\frac 1 t\\ &= \lim_{t\to 0^+}\frac{(2t+1)e^{-t}-1}{t}\\ &= \lim_{t\to 0^+}\frac{2e^{-t}-(2t + 1)e^{-t}}{1}\\ &= \lim_{t\to 0^+}(1-2t)e^{-t} = 1 \end{align}

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  • $\begingroup$ @oren revenge, seems to me you might have forgotten the minus sign when applying l'Hospital which would give your answer. $\endgroup$ – Ennar Mar 26 '16 at 12:15

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