Evaluation of limit at negative infinity is wrong

Question

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

I tried to raise $x$ to the square and get:

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

afterwards, as I divide the numerator and denominator by $x^2$ I get

$$\sqrt{\frac{1}{1+\frac1{x^2}}}$$

now after I set $x$ to minus infinity I get $1$. The answer is wrong though.It should be $-1$ as written in the answers.

Any guidance on how to solve it or directions to where I am mistaken will be appreciated.

Answer 1

$$\lim \limits_{x \to -\infty} \frac{x}{\sqrt{(x^2)+1}}$$

I tried to raise x to the square and get:

$$\lim \limits_{x \to -\infty} \sqrt\frac{x^2}{{(x^2)+1}}$$

So basically you replaced $x$ by $\sqrt{x^2}$ but these two expressions are only equal if $x$ is positive; $\sqrt{x^2}$ is positive (as it is the result of a square root) so for negative $x$, it cannot be equal to $x$.

You have $\sqrt{x^2}=|x|$ so for $x<0$, you have $\sqrt{x^2}=-x$ and you could replace $x$ by $-\sqrt{x^2}$.

Answer 2

You may write $$ \lim \limits_{x \to -\infty} \frac{x}{\sqrt{x^2+1}}=\lim \limits_{x \to -\infty} \left(\frac{x}{|x|}\cdot\sqrt{\frac{1}{1+\frac{1}{x^2}}}\right) $$ then observe that, as $x<0$, we have $$ \frac{x}{|x|}=-1. $$

Answer 3

The best way in these cases to avoid mistakes with sign or to check the result is change $x=-y\to \infty$ then

$$\lim \limits_{x \to -\infty} \frac{x}{\sqrt{x^2+1}}=\lim \limits_{y \to \infty} -\frac{y}{\sqrt{y^2+1}}=-1 $$