MHB Limit of a function when x goes to -infinity

[Yankel]

Hello all,

I have a small question. I was trying to graph this function:

\[\frac{x}{\sqrt{x^{2}+2}}\]

I have calculated it's limit when x goes to infinity, and got 1. I tried the same when it goes to minus infinity, and still got 1, because of the square. The answer should be -1, I don't understand why.

Can you assist ? Thank you !

 

[kaliprasad]

Hello all,

I have a small question. I was trying to graph this function:

\[\frac{x}{\sqrt{x^{2}+2}}\]

I have calculated it's limit when x goes to infinity, and got 1. I tried the same when it goes to minus infinity, and still got 1, because of the square. The answer should be -1, I don't understand why.

Can you assist ? Thank you !

as you know $\sqrt{x^2} > 0$for x < 0 numerator < 0 and denominator > 0 hence ratio = -ve

 

[mathmari]

Hello all,

I have a small question. I was trying to graph this function:

\[\frac{x}{\sqrt{x^{2}+2}}\]

I have calculated it's limit when x goes to infinity, and got 1. I tried the same when it goes to minus infinity, and still got 1, because of the square. The answer should be -1, I don't understand why.

Can you assist ? Thank you !

Hello!

$$\lim_{x \rightarrow +\infty } \frac{x}{\sqrt{x^2+2}}=\lim_{x \rightarrow +\infty } \frac{x}{\sqrt{x^2\left ( 1+\frac{2}{x^2} \right ) }}=
\lim_{x \rightarrow +\infty } \frac{x}{|x|\sqrt{1+\frac{2}{x^2} }}=(*)$$

While $x$ goes to $+\infty$, it is positive, so $|x|=x$.

$$(*)=\lim_{x \rightarrow +\infty } \frac{x}{x\sqrt{1+\frac{2}{x^2} }}=\lim_{x \rightarrow +\infty } \frac{1}{\sqrt{1+\frac{2}{x^2} }}=\frac{1}{\sqrt{1+0}}=1$$
$$\lim_{x \rightarrow -\infty } \frac{x}{\sqrt{x^2+2}}=\lim_{x \rightarrow -\infty } \frac{x}{\sqrt{x^2\left ( 1+\frac{2}{x^2} \right ) }}=
\lim_{x \rightarrow -\infty } \frac{x}{|x|\sqrt{1+\frac{2}{x^2} }}=(**)$$

While $x$ goes to $-\infty$, it is negative, so $|x|=-x$.

$$(**)=\lim_{x \rightarrow -\infty } \frac{x}{-x\sqrt{1+\frac{2}{x^2} }}=\lim_{x \rightarrow -\infty } \frac{1}{-\sqrt{1+\frac{2}{x^2} }}=-\frac{1}{\sqrt{1+0}}=-1$$