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 $1$?

  • $\begingroup$ 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. $\endgroup$
    – zipirovich
    Apr 25, 2021 at 5:37

2 Answers 2


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}

Consistent with Piskunov's answer.


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