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Find the limit $$ \lim_{x \to 0^+}{\sin^2x \over e^{-1/x}} $$

Have tried to apply the L'Hospital's rule, but the denominator stays always the same $\left( {d \over dx}e^x=e^x \right)$ and brings $-\infty$.

What else can be done if this powerful rule ships no help?

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3 Answers 3

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Change variable: as $x\rightarrow0^+$, then $-\frac1x\rightarrow-\infty.$ Hence $$ \lim_{x\rightarrow0^+}\frac{\sin^2x}{e^{-\frac1x}}\stackrel{-\frac1x=t}{=}\lim_{t\rightarrow-\infty}\frac{\sin^2(-1/t)}{e^t} $$

Observe now that $\sin^2(-1/t)\sim_{t\rightarrow-\infty}1/t^2$ thus $$ \lim_{t\rightarrow-\infty}\frac{\sin^2(-1/t)}{e^t}= \lim_{t\rightarrow-\infty}\frac{1}{t^2e^t}=+\infty $$ since $t^2e^t\rightarrow0^+$ as $t\rightarrow-\infty$.

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Using some Taylor polynomials

$$\sin^2x\cdot e^{1/x}=\sin^2x\left(1+\frac1x+\frac1{2x^2}+\frac1{3!x^3}+\ldots\right)\xrightarrow[x\to 0^+]{}\infty$$

since

$$\frac{\sin^x}{x^n}=\left(\frac{\sin x}x\right)^2\frac1{x^{n-2}}\xrightarrow[x\to 0]{}\infty\;\;\;\text{for}\;\;n>2$$

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  • $\begingroup$ btw Congrats @DonAntonio for reaching +100k !! $\overset{\cdot\cdot}\smile$ $\endgroup$
    – Hakim
    Commented Jun 8, 2014 at 12:02
  • $\begingroup$ Yeah I agree! Congrats DonAntonio!! $\endgroup$
    – Joe
    Commented Jun 8, 2014 at 12:22
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Be careful that :

$\frac{d}{dx}e^x=e^x$ but $\frac{d}{dx}e^{f(x)}\neq e^{f(x)}$

This should help you if you really understand what i have written...

Seeing $\lim _{x\rightarrow 0}\frac{\sin x}{x}= ??$ would be of some help to you...

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  • $\begingroup$ I know that ${d \over dx}e^{f(x)}=f^\prime(x)\cdot e^{f(x)}$ $\endgroup$ Commented Jun 8, 2014 at 11:23
  • $\begingroup$ So, where did you used it??? $\endgroup$
    – user87543
    Commented Jun 8, 2014 at 11:23
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    $\begingroup$ I derived numerator and denominator using L'Hospital's rule, which led me to $$\lim_{x \to 0+}{\sin2x \cdot x^2 \over e^{-1/x}}$$. I can apply it again but it won't help me to get rid of $e^{-1/x}$ in denominator. $\endgroup$ Commented Jun 8, 2014 at 11:26
  • $\begingroup$ I was trying to say (vaguely) what حكيم الفيلسوف الضائع has said in his comment.... do you understand what he said.. $\endgroup$
    – user87543
    Commented Jun 8, 2014 at 11:37

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