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 $1$?
 A: 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.
A: 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.
