# How do you find lim cosx/x as x->0^+?

Jun 17, 2017

$L {t}_{x \to {0}^{+}} \cos \frac{x}{x} = \infty$

#### Explanation:

In $L {t}_{x \to {0}^{+}} \cos \frac{x}{x}$, as $x \to 0$, $\cos x \to \cos 0 = 1$ and $x \to 0$

Hence $L {t}_{x \to {0}^{+}} \cos \frac{x}{x} = \frac{1}{0} = \infty$

Jun 21, 2017

${\lim}_{x \to {0}^{+}} \cos \frac{x}{x} = + \infty$

#### Explanation:

Apart from using the method shown by the other contributor, which is just plugging in 0 and finding that it approaches $\infty$, there is another, more sophisticated method of showing it, which is to use the Taylor approximation of $\cos x$ as $x \to 0$, or otherwise known as the Maclaurin expansion of $\cos x$.

The Maclaurin expansion of $\cos x$ is 1-x^2/2+x^4/(4!)-x^6/(6!)+...

Plugging in the Maclaurin expansion into the limit gives:

lim_(x->0^+)cosx/x=lim_(x->0^+)(1-x^2/2+x^4/(4!)-x^6/(6!)+...)/x

Simplifying gives:

lim_(x->0^+)1/x-x/2+x^3/(4!)-x^5/(6!)+...

When $x$ tends to 0, all the terms from the 2nd onwards become 0.
Therefore, the only term left is the first term, which is ${\lim}_{x \to {0}^{+}} \frac{1}{x}$. This leaves us with $+ \infty$, hence
${\lim}_{x \to {0}^{+}} \cos \frac{x}{x} = + \infty$