# 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...

• I have edited your post to make it more readable. Is this what you intended? – Cameron Williams Dec 4 '14 at 17:45
• exactly! it is difficult for me... – Pudding Dec 4 '14 at 17:48
• @Pudding If it's difficult for your, you might be interested in this basic information about writing math at this site: here, here, here and here. – Alice Ryhl Dec 4 '14 at 18:46
• @Pudding You may be interested in how to accept an answer. – AlexR Dec 4 '14 at 19:57

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 $L =\, ?$

• if $t=1$ then $\ln(1-t^2)$ is undefined. – Alice Ryhl Dec 4 '14 at 18:50
• @KristofferRyhl But it is of the form $\frac0{-\infty}$ wich is determinate as a limit. – AlexR Dec 4 '14 at 19:54
• Right, of course. – Alice Ryhl Dec 4 '14 at 19:55

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

$$\ln y = \frac{1}{2}\lim_{x\to 0}\frac{\ln(\cos x)}{\ln(\sin x)}.$$

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).