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Is it possible to find $$\lim_{x\to0}\frac{\sin(1-\cos(x))}{x^2e^x}$$ without using L'Hopital's Rule or Series or anything complex but just basic knowledge (definition of a limit, limit laws, and algebraic expansion / cancelling?)

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Multiply and divide the numerator and denominator by $1-\cos(x)$.

Then just basic trigonometric limits to evaluate the answer.

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  • $\begingroup$ Can you cancel the 1-cos(x) when sin of it is being taken? I wouldn't think it would cancel without dividing by sin(1-cos(x)) $\endgroup$ – user155301 Jun 5 '14 at 6:44
  • $\begingroup$ @user: No, you can't. You're correct to doubt it. $\endgroup$ – davidlowryduda Jun 5 '14 at 6:45
  • $\begingroup$ Yeah, sorry. To specify it isn't sinx(1-cos(x)) although that is what I thought when I first looked at it as well, it's sin(1-cos(x)). $\endgroup$ – user155301 Jun 5 '14 at 6:46
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    $\begingroup$ "Multiply and divide"? That would bring each of them back to their original form wouldn't it? $\endgroup$ – Marc van Leeuwen Jun 5 '14 at 11:49
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\begin{align} \lim_{x \to 0} \frac{\sin(1 - \cos x)}{\mathrm{e}^x \cdot x^2} &= \lim_{x \to 0} \frac{\sin(1 - \cos x)}{1 - \cos x} \cdot \frac{1 - \cos x}{x^2} \cdot \mathrm{e}^{- x} \\ &= \lim_{u \to 0} \frac{\sin u}{u} \cdot \lim_{x \to 0} \frac{1 - \cos x}{x^2} \cdot \lim_{x \to 0} \mathrm{e}^{- x} \\ &= 1\cdot \lim_{x \to 0} \frac{2 \sin^2 \frac{x}{2}}{x^2} \cdot 1 \\ &= \lim_{u \to 0} \frac{2 \sin^2 u}{4 u^2} \end{align}

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  • $\begingroup$ There's still a limit as x goes to 0 of something over x^2 which makes it not work.. $\endgroup$ – user155301 Jun 5 '14 at 9:14
  • $\begingroup$ @user155301: I dont get what issue you face with $x^{2}$ $\endgroup$ – Paramanand Singh Jun 5 '14 at 9:15
  • $\begingroup$ @user155301 even if your proof can't use series, look at the series for $\cos x$... $\endgroup$ – vonbrand Jun 5 '14 at 9:16
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    $\begingroup$ we have $(1 - \cos x)/x^{2} = 2\sin^{2}(x/2)/x^{2} = 2\sin^{2}(x/2)/(x/2)^{2}\cdot(x/2)^{2}/x^{2} \to 2/4 = 1/2$ $\endgroup$ – Paramanand Singh Jun 5 '14 at 9:17
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    $\begingroup$ @vonbrand: i don't see any issue with $1 - \cos x = 2\sin^{2}(x/2)$ $\endgroup$ – Paramanand Singh Jun 5 '14 at 9:21
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We can proceed as follows: $$\begin{aligned}L &= \lim_{x \to 0}\frac{\sin(1 - \cos x)}{x^{2}e^{x}}\\ &= \lim_{x \to 0}\frac{\sin(1 - \cos x)}{1 - \cos x}\cdot\frac{1 - \cos x}{x^{2}e^{x}}\\ &= \lim_{x \to 0}\frac{\sin(1 - \cos x)}{1 - \cos x}\cdot\frac{1 - \cos x}{x^{2}}\cdot \frac{1}{e^{x}}\\ &= \lim_{t \to 0}\frac{\sin t}{t}\cdot\lim_{x \to 0}\frac{1 - \cos x}{x^{2}}\cdot \lim_{x \to 0}\frac{1}{e^{x}}\text{ by putting }t = 1 - \cos x \to 0 \text{ as } x \to 0\\ &= 1\cdot\lim_{x \to 0}\frac{1 - \cos x}{x^{2}}\cdot 1\\ &= \lim_{x \to 0}\frac{2\sin^{2}(x/2)}{(x/2)^{2}}\cdot\frac{(x/2)^{2}}{x^{2}}\\ &= 2\cdot 1\cdot\frac{1}{4} = \frac{1}{2}\end{aligned}$$

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  • $\begingroup$ I delete my answer to avoid irregular downvotes but how did you explain the other answers on this post? It seems they all (except yours) used the same method like I used. I really appreciate if you want to discuss this issue to me. $\endgroup$ – Tunk-Fey Jun 5 '14 at 10:34
  • $\begingroup$ @Tunk-Fey: All the other answers (except mine) in that linked post work on this theorem: if $f(x) \to A$ and $g(x) \to B$ and $A > 0$ then $\{f(x)\}^{g(x)} \to A^{B}$. $\endgroup$ – Paramanand Singh Jun 5 '14 at 10:40
  • $\begingroup$ OK, I can accept your argument. Thanks. $\endgroup$ – Tunk-Fey Jun 5 '14 at 10:55
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    $\begingroup$ @Tunk-Fey: by the way thanks for the discussion. You may visit my blog post paramanands.blogspot.com/2013/11/… for more details (especially the section on "Misuse of Rules of Limits" $\endgroup$ – Paramanand Singh Jun 5 '14 at 11:06
  • $\begingroup$ OK, +1 for your answer. $\endgroup$ – Tunk-Fey Jun 5 '14 at 12:18

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