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I am trying to evaluate this limit:

$\lim_{x \to 0} \dfrac{(\sin x -x)(\cos x -1)}{x(e^x-1)^4}$

I know that I need to use Taylor expansions for $\sin x -x$, $\cos x -1$ and $e^x-1$. I also realise that all of these are just their regular Taylor expansions with their first term removed so the series starts at $n=1$ rather than at 0.

However, when I actually try to evaluate I get stuck at:

$\lim_{x \to 0} \dfrac{(-\frac{1}{3!}+\frac{x^2}{5!}...)(-\frac{9x^2}{2!}+\frac{81x^4}{4!}...)}{(\frac{1}{x}+\frac{1}{2!}+\frac{x}{3!}+\frac{x^2}{4!}...)^4}$

I don't know how to proceed from here.

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  • $\begingroup$ Not sure how you got your terms... I think the leading term in each case works out so that $x$ has equal lowest exponent on top and bottom... $\endgroup$ – abiessu Apr 21 '16 at 3:20
  • $\begingroup$ I just factored $x^3$ out of $\sin x - x$ and out of the denominator. $\endgroup$ – jessicajjensen Apr 21 '16 at 3:26
  • $\begingroup$ Wow, I just realized that when I factored I completely ignored the $()^4$ on the denominator, let me redo my computations. $\endgroup$ – jessicajjensen Apr 21 '16 at 3:26
  • $\begingroup$ Wow. Solution was incredibly simple. I feel so foolish now. $\endgroup$ – jessicajjensen Apr 21 '16 at 3:30
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    $\begingroup$ No worries, you solved it! $\endgroup$ – abiessu Apr 21 '16 at 3:33
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Just in case you wanted something to check your answer off of:

\begin{align} \lim_{x \to 0} \dfrac{(\sin x -x)(\cos x -1)}{x(e^x-1)^4} &= \lim_{x\to 0} \frac{\left(\frac{-x^3}{3!}+\frac{x^5}{5!} - \cdots\right)\left(\frac{-x^2}{2!} + \frac{x^4}{4!}- \cdots\right)}{x\left(x+\frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\right)^4}\\ &= \lim_{x\to 0} \frac{x^3\left(\frac{-1}{3!}+\frac{x^2}{5!} - \cdots\right)x^2\left(\frac{-1}{2!} + \frac{x^2}{4!}- \cdots\right)}{x^5\left(1+\frac{x}{2!} + \frac{x^2}{3!} + \cdots\right)^4}\\ &= \lim_{x\to 0} \frac{\left(\frac{-1}{3!}+\frac{x^2}{5!} - \cdots\right)\left(\frac{-1}{2!} + \frac{x^2}{4!}- \cdots\right)}{\left(1+\frac{x}{2!} + \frac{x^2}{3!} + \cdots\right)^4}\\ &= \left(\frac{-1}{3!}\right)\left(\frac{-1}{2!}\right)\\ &= \frac{1}{12} \end{align}

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  • $\begingroup$ After graphing in DESMOS, your answer seems to be correct indeed! $\endgroup$ – imranfat Apr 21 '16 at 3:44
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    $\begingroup$ I checked it using a couple of other techniques before I posted it. I have also edited it so that it's a bit clearer what I did to get there. $\endgroup$ – Nicholas Stull Apr 21 '16 at 3:47
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Since we don't need Taylor Series at all, I thought it might be instructive to present a way forward that avoids series expansions.

Note that we can write

$$\begin{align} \frac{(\sin(x)-x)(\cos(x)-1)}{x(e^x-1)^4}&=-\frac{\frac{\sin(x)-x}{x^3}\,\frac{1-\cos(x)}{x^2}}{\left(\frac{e^x-1}{x}\right)^4}\\\\ &-\frac{\frac{\sin(x)-x}{x^3}\,\frac{2\sin^2(x/2)}{x^2}}{\left(\frac{e^x-1}{x}\right)^4} \end{align}$$

Now, it is easy to show using the inequalities from elementary geometry

$$|x\cos(x)|\le |\sin(x)|\le |x|$$

for $|x|\le \pi/2$, that

$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 0}\frac{2\sin^2(x/2)}{x^2}=\frac12}$$

In addition, I showed in THIS ANSWER using only the limit definition of the exponential function and Bernoulli's Inequality that

$$1+x\le e^x\le \frac{1}{1-x}$$

for $x<1$. Then, it is easy to see that

$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 0}\left(\frac{e^x-1}{x}\right)^4=1}$$

We are left only to find the limit

$$\begin{align} \lim_{x\to 0}\frac{\sin(x)-x}{x^3}&=\lim_{x\to 0}\frac{\cos(x)-1}{3x^2}\\\\ &=-\frac16 \end{align}$$

using L'Hospital's Rule once followed by using the aforementioned limit $\lim_{x\to 0}\frac{2\sin^2(x/2)}{x^2}=\frac12 $.

Putting it all together, we find the limit of interest is

$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 0}\frac{(\sin(x)-x)(\cos(x)-1)}{x(e^x-1)^4}=\frac{1}{12}}$$

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  • $\begingroup$ A great alternative (+1) but if I show this approach to my calculus newbees, they are going to throw oranges at me...:) $\endgroup$ – imranfat Apr 21 '16 at 13:55
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    $\begingroup$ A very nice alternative to power series. @imranfat, if you're lucky, they'll pick something as soft as oranges (I've had a class joke about throwing a desk at me for working through a power series solution to an ODE). $\endgroup$ – Nicholas Stull Apr 21 '16 at 15:50
  • $\begingroup$ @imranfat Thank you! Just be sure to bring an orange juice maker to class ... oh and a baseball glove to catch the oranges. $\endgroup$ – Mark Viola Apr 21 '16 at 16:17
  • $\begingroup$ @nicholasstull Thank you! Your class must be pretty strong to be able to throw a desk. $\endgroup$ – Mark Viola Apr 21 '16 at 16:18
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Turns out to just be a foolish mistake. When I factored $x$ out of the denominator I should have raised it to 4. That would be an $x^5$ on the bottom:

$\lim_{x \to 0} \dfrac{x^3(-\frac{1}{3!}+\frac{x^2}{5!}...)x^2(-\frac{9}{2!}+\frac{81x^2}{4!}...)}{x^5({1}+\frac{x}{2!}+\frac{x^2}{3!}+\frac{x^3}{4!}...)^4}$

Now I can simply cancel out the $x^5$s and then substitute remaining $x$s for zeroes.

$\lim_{x \to 0} \dfrac{(-\frac{1}{3!})(-\frac{9}{2!})}{({1})^4} = \dfrac{3}{4} $

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  • $\begingroup$ Why do you have the leading term on $(\cos(x)-1)$ as $-\frac{9}{2!}x^2$? It should just be $-\frac{x^2}{2!}$, since $$\cos(x)=1-\frac{x^2}{2!} + \frac{x^4}{4!} - \cdots$$ $\endgroup$ – Nicholas Stull Apr 21 '16 at 3:40
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    $\begingroup$ @jessicajjensen. Something is not right with your computation. When graphing the function, it seems like the limit concurs with Nicholas' answer... $\endgroup$ – imranfat Apr 21 '16 at 3:45
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    $\begingroup$ @imranfat, the computation has a problem right in that first line, in the second part of the numerator (the stuff in parentheses after $x^2$, where she computed $\cos(x)-1$). It should just be $$\left(\frac{-1}{2!} + \frac{x^2}{4!} - \cdots\right)$$ Her expansion would seem to be for $\cos(9x)-1$, which would have led to her answer if that was indeed the problem. $\endgroup$ – Nicholas Stull Apr 21 '16 at 3:50
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Using basic limits \begin{eqnarray*} \lim_{x\rightarrow 0}\left( \frac{\sin x-x}{x^{3}}\right) &=&-\frac{1}{6} \\ \lim_{x\rightarrow 0}\left( \frac{\cos x-1}{x^{2}}\right) &=&-\frac{1}{2} \\ \lim_{x\rightarrow 0}\left( \frac{x}{e^{x}-1}\right) &=&1 \end{eqnarray*} which can be evaluated using L'HR, it follows that \begin{eqnarray*} \lim_{x\rightarrow 0}\frac{(\sin x-x)(\cos x-1)}{x(e^{x}-1)^{4}} &=&\lim_{x\rightarrow 0}\left( \frac{\sin x-x}{x^{3}}\right) \left( \frac{% \cos x-1}{x^{2}}\right) \left( \frac{x}{e^{x}-1}\right) ^{4} \\ &=&\left( \lim_{x\rightarrow 0}\frac{\sin x-x}{x^{3}}\right) \left( \lim_{x\rightarrow 0}\frac{\cos x-1}{x^{2}}\right) \left( \lim_{x\rightarrow 0}\frac{x}{e^{x}-1}\right) ^{4} \\ &=&\left( -\frac{1}{6}\right) \left( -\frac{1}{2}\right) \left( 1\right) ^{4} \\ &=&\frac{1}{12}. \end{eqnarray*}

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