135. Find $\lim _{x \rightarrow \infty} \frac{\sqrt{x^{2}+4}}{x}$.
Solution. As the argument $x$ tends to infinity, we have an indeterminate form of $\infty / \infty$. To resolve it, we divide the numerator and the denominator of the fraction by $x$. Then we get

$$
\lim _{x \rightarrow \infty} \frac{\sqrt{x^{2}+4}}{x}=\lim _{x \rightarrow \infty} \frac{\sqrt{\frac{x^{2}+4}{x^{2}}}}{1}=\lim _{x \rightarrow \infty} \sqrt{1+\frac{4}{x^{2}}}=1
$$

since $4 / x^{2} \rightarrow 0$ as $x \rightarrow \infty$.