Evaluating $\lim_{x \rightarrow 0}(1-\cos x)^{\tan x}$

Question

I have the following limit to solve.

$$\lim_{x \rightarrow 0}(1-\cos x)^{\tan x}$$

I am normally supposed to solve it without using l'Hôpital, but I failed to do so even with l'Hôpital. I don't see how I can solve it without applying l'Hôpital a couple of times, which doesn't seem practical, nor how to solve the question without applying it. Thanks for the help.

Answer 1

Note that

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

Now, since $\lim_{x\to 0}2^{\tan(x)}=1$, $\lim_{x\to 0}4\cos(x/2)/\cos(x)=4$, and $\lim_{x\to 0}x^x=1$, then $\lim_{x\to 0}(\sin(x/2))^{\sin(x/2)}=1$, and the limit of interest is $1$.

Answer 2

Try with Taylor series when $x\to0$: $$\tan{x}\approx x$$ $$\cos{x}\approx 1-\frac{x^2}{2}$$

Answer 3

Let $y=\lim_{x\to0}(1-\cos x)^{\tan x}$

$$\ln y=\lim_{x\to0}\dfrac{\ln(1-\cos x)}{\cot x}$$ which is of the form $\dfrac\infty\infty$

So applying L'Hospital $$\ln y=-\lim_{x\to0}\dfrac{\sin x}{\csc^2x(1-\cos x)} =-\lim_{x\to0}\dfrac{\sin x(1+\cos x)(1-\cos x)}{(1-\cos x)}=?$$

Answer 4

Doing this without L'hopital is tricky, but you have to make estimates.

If the know the graph of $cos(x)/tan(x)$, $cos(0) = 1$, and $tan(0) = 0$

So you have:

$$(1- cos(x))^{tan(x)}$$ $$(0)^{0}$$

But if we estimate (right hand side limit), we have something like:

$$(1- 0.01)^{0.01}$$ $$(0.09)^{0.01}$$ Which is approx 1.

Answer 5

Even if, apparently, Taylor series are not desired, may be equivalents could be used $$A=(1-\cos(x))^{\tan(x)}\implies \log(A)=\tan(x) \log(1-\cos(x))$$ Close to $x=0$, $$\cos(x)\sim 1-\frac{x^2} 2$$ $$1-\cos(x)\sim \frac{x^2} 2$$ $$\log(1-\cos(x))\sim 2\log(x)-\log(2)$$ $$\tan(x)\sim x$$ $$log(A)=\tan(x) \log(1-\cos(x))\sim 2x\log(x)-x\log(2)$$ Now, when $x\to 0$, each of the two terms goes to $0$; so $\log(A)\to 0$ and then $A\to 1$.