L'Hospital's rule help

I tried using L'Hospital's rule to find the limit as $x$ tends to $0$ for the following function:

$$f(x) = \frac{(1 - \cos x)^{1.5}}{x - \sin x}$$ I tried differentiating the top and the bottom and I can do it many times but it still gives the denominator with zero meaning I have to differentiate again and again (mainly because a $(1 - \cos x)^{0.5}$ always manages to find its way into the numerator.)

Is there a better way of solving this limit question?

• You won't have to use Hospital more than 3 times. After the third time, your denominator becomes $cosx$ and so $cos0=1$. So be a little persistent... – imranfat Apr 19 '17 at 1:48
• no because then in the numerator i get a fraction which has a root (1-cos(x) ) in its denominator so that gives me problems.... and then its just a vicious cycle.... – David Abraham Apr 19 '17 at 1:53
• use a substitution – user29418 Apr 19 '17 at 1:57
• @DavidAbraham. Actually you are right. So I proposed a different (and quite easy) answer if you are familiar with conjugates. Also, make a graph and see what is going on at $x=0$... – imranfat Apr 19 '17 at 2:26
• As a check, for small $x$ you have $\cos x \approx 1 -\dfrac{x^2}{2}$ and $\sin x \approx x - \dfrac{x^3}{6}$ so $f(x)\approx \dfrac{\left( \frac{x^2}{2}\right)^{3/2}}{ \frac{x^3}{6}} = \dfrac{3}{\sqrt{2}} \dfrac{|x|^3}{x^3}$ which is non-zero but changes sign around $x=0$ – Henry Apr 19 '17 at 7:59

Hint:

$$1-\cos x = 2\sin^2 \frac{x}{2}$$

also:

$$\frac{2\sin \frac{x}{2}}{x - 2\sin \frac{x}{2}\cos \frac{x}{2}}$$

$$\implies \frac{2}{\frac{1}{2} \frac{\frac{x}{2}}{\sin \frac{x}{2}} -{2}\cos \frac{x}{2}}$$

where:

$$\lim_{\alpha \rightarrow 0} \frac{\sin \alpha}{\alpha} = 1$$

here consider: $\alpha = \frac{x}{2}.$

Then, it's ready for replacing $x \rightarrow 0$!

• Thanks this helped a lot! – David Abraham Apr 19 '17 at 1:58
• You're welcome. – Amin Apr 19 '17 at 2:00

It appears that you mean $(1 - \cos x)^{3/2}$ in the numerator. Please reflect this by editing your question.I have edited the question to reflect this.

In this case it can be shown that the limit does not exist as left and right hand limits are different. Doing this requires the use of standard limit $$\lim_{x \to 0}\frac{1 - \cos x}{x^{2}} = \frac{1}{2}\tag{2}$$ and the use of L'Hospital's Rule once.

Thus we have for $x \to 0^{+}$ \begin{align} L &= \lim_{x \to 0^{+}}\frac{(1 - \cos x)^{3/2}}{x - \sin x}\notag\\ &= \lim_{x \to 0^{+}}\left(\frac{1 - \cos x}{x^{2}}\right)^{3/2}\cdot\frac{x^{3}}{x - \sin x}\notag\\ &= \frac{1}{2\sqrt{2}}\lim_{x \to 0^{+}}\frac{x^{3}}{x - \sin x}\notag\\ &= \frac{1}{2\sqrt{2}}\lim_{x \to 0^{+}}\frac{3x^{2}}{1 - \cos x}\text{ (via L'Hospital's Rule)}\notag\\ &= \frac{3}{\sqrt{2}}\notag \end{align} The limit as $x \to 0^{-}$ is $-3/\sqrt{2}$ and this can be shown by putting $x = -t$ and then using the above limit evaluation.

Note that a two sided limit does not exist, so let's look at the limit approaching from the positive side. After one time Hospital, we get $\frac{1.5sinx\sqrt{1-cos}}{1-cosx}$.Now it goes very easy: Divide out $\sqrt{1-cosx}$ and then multiply top and bottom by $\sqrt{1+cosx}$ and after applying the Pythagorean theorem in the denominator, the expression becomes $\frac{1.5sinx\sqrt{1+cosx}}{sinx}$. So the $sinx$ term cancels and you can take the limit $\frac{3\sqrt{2}}{2}$. The reason why the limit only exists one sided is because technically the squareroot of $sin^2x$ is $|sinx|$. Hint: Make a graph. I did just that...When you approach $0$ from the negative side, you get $-\frac{3\sqrt{2}}{2}$. Do you see why?

Note that $1-\cos x\geq 0$ and is an even function, so $(1-\cos x)^{3/2}$ is also even. On the other hand, $x-\sin x$ is odd. When you divide an even and odd function, you get an odd function. Due to the symmetry of odd functions, either $\lim_{x\rightarrow 0} f(x) = 0$ or the limit doesn't exist. You can see that the limit on one side is $\frac{3}{\sqrt{2}}$ (see imranfat's answer), so the limit must not exist.