Help me please limit without L'Hospital

Question

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...

Answer 1

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?

Answer 2

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 $L =\, ?$

Answer 3

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).