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How to work around to find the limit for these functions :

  1. $$\lim_{h\rightarrow 0} \frac{e^{-h}}{-h}$$

  2. $$\lim_{h\rightarrow 0} \frac{|\cos h-1|}{h}$$

For the second one i think that the limit doesn't exist.

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  • $\begingroup$ Do you know the Taylor series for $\cos$? $\endgroup$ – Daniel Fischer Jul 6 '13 at 15:31
  • $\begingroup$ for 1st one $e^{-h}$ will be always one but 1/-h will depend upon left and right hand limit $\endgroup$ – Bhauryal Jul 6 '13 at 15:36
  • $\begingroup$ Since you've asked the other as a new question already and have several fine answers to your original question, I have rolled your question back. (You risked the second being closed as a duplicate and then getting no answers on it.) $\endgroup$ – Cameron Buie Jul 6 '13 at 16:25
  • $\begingroup$ @CameronBuie Thanks !! $\endgroup$ – Aman Mittal Jul 6 '13 at 16:28
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HINT:

$(1):\lim_{h\to0}e^{-h}=1$

$(2):$ $$\cos h=1-2\sin^2\frac h2\implies \cos h-1=-2\sin^2\frac h2$$

$$\implies \frac{\cos h-1}h=-\left(\frac{\sin \frac h2}{\frac h2}\right)^2 \frac h4$$

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  • $\begingroup$ Thanks !! I should have thought about this method for the second question .. $\endgroup$ – Aman Mittal Jul 6 '13 at 15:56
  • $\begingroup$ @AmanMittal, my pleasure. It's always useful to memorize Trigonometric Identities while solving the problems on limit. Hope I could make the idea clear $\endgroup$ – lab bhattacharjee Jul 6 '13 at 15:57
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Note that $\cos h\le 1$ so $|\cos h-1|=1-\cos h$.

$$\lim_{h\to 0} \frac{1-\cos h}{h}=\lim_{h\to 0} \frac{(1-\cos h)(1+\cos h)}{h(1+\cos h)}=\lim_{h\to 0} \frac{1-\cos^2 h}{h(1+\cos h)}=\lim_{h\to 0}\frac{\sin h}{h}\lim_{h\to 0}\frac{\sin h}{1+\cos h}=1\cdot 0=0$$

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The first one doesn't exist, because $\lim_{h\to 0}e^{-h}=1$. The second one can be done either via Taylor series, l'Hopital's rule or the the following trick:

$$\frac{\cos h-1}{h}=\frac{\cos h - \cos 0}{h-0}\to (\cos h )'|_{h=0}=0.$$

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  • 3
    $\begingroup$ Your "trick" is a bit circular... $\endgroup$ – David Mitra Jul 6 '13 at 15:43
  • $\begingroup$ @DavidMitra what do you mean? $\endgroup$ – TZakrevskiy Jul 6 '13 at 15:52
  • $\begingroup$ I mean, in many treatments, the limit you're trying to compute is used to show that the derivative of $\cos x$ is $-\sin x$ (I assume that's what you used in the final equality). $\endgroup$ – David Mitra Jul 6 '13 at 15:56
  • $\begingroup$ @DavidMitra I see. $\endgroup$ – TZakrevskiy Jul 6 '13 at 16:02
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Use the reverse Prosthaphaeretic identities.

$$\cos(a)-\cos(b)=-2\sin((a+b)/2)\sin((a-b)/2)$$

Taking into account that $\cos(0)=1$

Then you can write

$$\frac{\cosh-1}{h}=\frac{-2\sin(h/2)\sin(h/2)}{h}.$$

Use then that

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We know that $e^x=\frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots$ Note that for $|h|<1$ and sufficiently small, $$ \frac{1}{0!}(-h)^0+\frac{1}{1!}(-h)^1 \leq e^{-h} \leq \frac{1}{0!}(-h)^0+\frac{1}{1!}(-h)^1+\frac{1}{2!}(-h)^2 $$ which is the same as $$ 1-h \leq e^{-h} \leq 1-h+\frac{1}{2}h^2 $$ implies $$ 1-\frac{1}{h} \geq \frac{e^{-h}}{-h} \geq 1-\frac{1}{h}-\frac{1}{2}h $$ Make $h\to 0$. What you get? Now the other limit. Note that \begin{align} \frac{|\cos(h)-1|}{h} = & 2\frac{\left|\cos\left(2\dfrac{h}{2}\right)-1\right|}{\dfrac{h}{2}} \\ = & 2\frac{\left|1-2\sin^2\left(\dfrac{h}{2}\right)-1\right|}{\left(\dfrac{h}{2}\right)} \\ = & 4\frac{\left|\sin^2\left(\dfrac{h}{2}\right)\right|}{\left(\dfrac{h}{2}\right)} \\ = & 4\left(\dfrac{h}{2}\right)\frac{\sin^2\left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)^2} \end{align} Make $h\to 0$. What you get?

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  • $\begingroup$ Just to clarify. The intent of these algebraic manipulations is to use the fact that $\lim_{h\to 0}\frac{\sin \frac{h}{2}}{\frac{h}{2}}=1$, $\lim_{h\to 0}\frac{h}{2}=0$, $\lim_{h\to 0}1-\frac{1}{h}=-\infty$ and $\lim_{h\to 0}1-\frac{1}{h}-\frac{1}{2}h=-\infty$ $\endgroup$ – MathOverview Jul 6 '13 at 16:30

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