# What is $\lim\limits_{x \to \infty}\dfrac{\sqrt{x^2+1}}{x+1}$?

This question is from Differential and Integral Calculus by Piskunov. I've to evaluate the following limit:

$$\lim\limits_{x \to \infty}\dfrac{\sqrt{x^2+1}}{x+1}$$

This is how I tried solving it,

Put $$t=\frac{1}{x}$$. Then the limit becomes

$$\lim\limits_{t \to 0}\dfrac{\sqrt{\dfrac{1}{t^2}+1}}{\dfrac{1}{t}+1}$$

Simplifying a bit gets me,

$$\lim\limits_{t \to 0}\dfrac{\sqrt{1+t^2}}{1+t}$$

As $$t \to 0$$, I think $$\dfrac{\sqrt{1+t^2}}{1+t} \to 1$$. But according to Piskunov, if $$x \to +\infty$$ then the limit is $$+1$$, and if $$x \to -\infty$$ then the limit is $$-1$$.

Why is the limit not simply $$1$$?

• It's not true, although it is a very common misconception, that for all real numbers "$\sqrt{t^2}=t$". The correct statement is that for all real numbers $\sqrt{t^2}=|t|$, to account for the case when $t$ is negative. Commented Apr 25, 2021 at 5:37

You've missed something in your simplification.

$$\frac{\sqrt{\frac1{t^2}+1}}{\frac1t+1} = \frac{t}{|t|}\frac{\sqrt{1+t^2}}{1+t} = \text{sgn}(t)\frac{\sqrt{1+t^2}}{1+t}$$

So for $$t\to 0^{-}$$ or $$x\to -\infty$$, you have $$\text{sgn}(t) = -1$$

and for $$t\to 0^{+}$$ or $$x\to +\infty$$, you have $$\text{sgn}(t) = 1$$.

This will yield the required result.

Given $$\lim\limits_{x \to +\infty}\dfrac{\sqrt{x^2+1}}{x+1}$$, set $$t=\frac{1}{x},x=\frac{1}{t} ; x \to +\infty, t\to \frac{1}{+\infty}\to 0^+$$

\begin{align}\boxed{\lim\limits_{x \to +\infty}\dfrac{\sqrt{x^2+1}}{x+1}=\lim\limits_{t \to 0^+}\dfrac{\sqrt{\dfrac{1}{t^2}+1}}{\dfrac{1}{t}+1}=\lim\limits_{t \to 0^+}\dfrac{\sqrt{1+t^2}}{1+t}=1}\end{align}

Given $$\lim\limits_{x \to -\infty}\dfrac{\sqrt{x^2+1}}{x+1}$$, set $$t=\frac{1}{x},x=\frac{1}{t} ; x \to -\infty, t\to \frac{1}{-\infty}\to 0^-$$

\begin{align}\boxed{\lim\limits_{x \to -\infty}\dfrac{\sqrt{x^2+1}}{x+1}=\lim\limits_{t \to 0^-}\dfrac{\sqrt{\dfrac{1}{t^2}+1}}{\dfrac{1}{t}+1}=\lim\limits_{t \to 0^-}\dfrac{-\sqrt{1+t^2}}{1+t}=-1}\end{align}