Help me please limit without L'Hospital I have a question:
How to calculate
$$\lim_{x\to 0}(\cos x)^{\dfrac{1}{\ln(\sin^2(x))}}$$
without L'Hospital rule...
Help me please...
 A: Hint
$\exp$ is  continuous, so your limit is equal to
$$L=\exp\left(\lim_{x\to 0} \frac{\ln(\cos x)}{\ln(\sin^2(x))}\right) = \exp\left(\lim_{x\to0} \frac{\ln(\cos x)}{\ln(1-\cos^2 x)}\right)$$
Now substitute $t = \cos x$ so you need to find
$$L=\exp\left(\lim_{t\to1} \frac{\ln t}{\ln(1-t^2)} \right)$$
Wich is no longer an indeterminate form. Conclude $L =\, ?$
A: Often when you have limits with powers like this, you want to use logarithms. Let
$$y = \lim_{x\to 0}(\cos x)^{\dfrac{1}{\ln(\sin^2(x))}},$$
then 
$$\ln y = \lim_{x\to 0}\frac{1}{\ln(\sin^2(x))}\ln(\cos x)$$
Since $\ln(\sin^2(x)) = \ln((\sin x)^2) = 2\ln(\sin x)$, we get
$$\ln y = \frac{1}{2}\lim_{x\to 0}\frac{\ln(\cos x)}{\ln(\sin x)}.$$
What do you know about this last limit? What then do you conclude about your original limit?
A: Use logarithmic differentiation. Basic idea: if you have $$y = [f(x)]^{g(x)}\text{,}$$ then $$\ln(y) = g(x)\ln[f(x)]$$
and as long as $x_0 > 0$, by continuity of $\ln$, 
$$\lim\limits_{x \to x_0}\ln(y) = \ln\left[\lim\limits_{x \to x_0}y\right]$$
and thus your problem ends up being 
$$\ln\left[\lim\limits_{x \to x_0}y\right] = \lim\limits_{x \to x_0}g(x)\ln[f(x)]\text{.}$$
Since $y$ is equivalent to the desired function, exponentiating on both sides, your problem reduces to
$$\lim\limits_{x \to x_0}y = \lim\limits_{x \to x_0}[f(x)]^{g(x)} =  e^{\lim\limits_{x \to x_0}g(x)\ln[f(x)]}\text{.}$$
So, 
$$\lim\limits_{x \to 0}\{[\cos(x)]^{1/\ln[\sin^{2}(x)]}\} = e^{\lim\limits_{x \to 0}\frac{\ln[\cos(x)]}{\ln[\sin^{2}(x)]}}\text{.}$$
Thus your challenge is to find
$$\lim\limits_{x \to 0}\dfrac{\ln[\cos(x)]}{\ln[\sin^{2}(x)]}\text{.}$$
Notice here that $$\lim\limits_{x \to 0}\cos(x) = 1$$
and 
$$\lim\limits_{x \to 0}\sin^{2}(x) = 0$$
so that, as $x \to 0$, $\ln[\cos(x)] \to \ln(1) = 0$ and by (right-)continuity of $\ln$, $\ln[\sin^{2}(x)] \to -\infty$ (visualize the graph of $\ln$ to see this).
