MHB Limit of a function when x goes to -infinity

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The limit of the function \(\frac{x}{\sqrt{x^{2}+2}}\) as \(x\) approaches infinity is correctly calculated as 1. However, when evaluating the limit as \(x\) approaches negative infinity, the absolute value of \(x\) must be considered, leading to a negative result. Specifically, as \(x\) goes to negative infinity, the limit simplifies to -1. This discrepancy arises from the sign of \(x\) affecting the ratio in the limit calculation. Understanding the behavior of the square root and the sign of \(x\) is crucial for accurate limit evaluation.
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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 !
 
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Yankel said:
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
 
Last edited:
Yankel said:
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 $|x|=-x$.

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