How can I solve this limit the easiest way?

Question

Find the limit of:

$\lim_\limits{x \to 0^+} \frac{\cos x}{\log x}$

Answer 1

First of all, note this limit doesnt exist, once log x is not defined for x < 0. Assuming you want to compute this limit with x -> 0+, the answer is quite simple. The cosine will tend to 1 and the log x will tend to negative infinity. That yields 0.

Answer 2

We have

$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 0^+}\frac{\cos(x)}{\log(x)}=0}$$

since for all $\epsilon>0$, $\left|\frac{\cos(x)}{\log(x)}\right|<\epsilon$ whenever $0<x<\min(1,e^{-1/\epsilon})$.

Alternatively, I showed in THIS ANSWER using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm satisfies the inequalities

$$\frac{x-1}{x}\le \log(x)\le x-1 \tag 1$$

for $x>0$.

Then, for $0<x<1$, we have using $(1)$

$$0\le \left|\frac{\cos(x)}{\log(x)}\right|<\frac{x}{1-x}$$

whereupon application of the squeeze theorem yields the coveted limit.

Answer 3

If you are working in $\mathbb{R}$ only the right handed limit will exist. In this case the we note that $0 < \frac{\cos x}{\log x} < \frac{1}{\log x}$. Taking the limit $0^+$ we find the answer to be $0$.

If we are working in $\mathbb{C}$ then we note that we can rewrite the limit as $\lim_{z\to 0}\frac{\cos z}{\log |z| + i\arg(z)}$. Through similar manipulations we find that the complex limit is indeed $-\infty$ as well