Outline: Let $P(a)=b\gt 0$. Then by the continuity of $P(x)$ at $x=a$, there exists a $\delta_1$ such that if $|x-a|\lt \delta_1$, then $|P(x)-b|\lt \frac{b}{2}$. In particular, if $|x-a|\lt \delta_1$, we have $P(x)\gt \frac{b}{2}$.
We want to show that for any $M\gt 0$, there exists a $\delta$ such that if $0\lt |x-a|\lt \delta$, then $\left|\frac{P(x)}{x-a}\right|\gt M$.
We will make sure later that $\delta\le \delta_1$. Then if $0\lt |x-a|\lt \delta$, we have $\frac{P(x)}{|x-a|}\gt \frac{b}{2\delta}$. To make this $\gt M$, it is enough to take $\delta\le \frac{b}{2M}$.
Putting things together, we want $\delta=\min\left(\delta_1, \frac{b}{2M}\right)$.
If $0\lt x-a\lt \delta$, then $\frac{P(x)}{x-a}$ is positive, and greater than $M$. It follows that
$$\lim_{x\to a^+}\frac{P(x)}{x-a}=\infty.$$
By the same inequalities, if $x\lt a$ and $|x-a|\lt \delta$, then $\frac{P(x)}{x-a}$ is less than $-M$. It follows that
$$\lim_{x\to a^-}\frac{P(x)}{x-a}=-\infty.$$
By definition, $\lim_{x\to\infty}f(x)=L$ if $\forall\epsilon\gt 0$ there exists $N\gt 0$ such that $x\gt N\implies |f(x)-L|\lt\epsilon$
$|f(x)-L|\lt\epsilon$ is equivalent to $L-\epsilon\lt f(x)\lt L+\epsilon$.
To show that the limit does not exist, we can show that for any $L$, there is a fixed $\epsilon$ for which the inequality cannot hold.
Since we are talking about the square root we can assume $L$ is positive. Let $\epsilon = 1$. Then it is sufficient to show that we can always find $x$ such that $\sqrt{x}\ge L+1$.
Let $x=max(N+1, (L+2)^2)$. Then $x\gt N$ and $\sqrt{x}=\sqrt{(L+2)^2}=L+2\gt L+1$ so $L$ cannot be the limit. Since this is true for all $L$, the square root does not converge to a limit as $x$ approaches infinity.
Best Answer
Hint: Note that $\cos x=\cosh(ix)$. Now what happens to $\cos x$ as $x\to\infty$ along the $x$-axis?