Limit of a function when x goes to -infinity

  • Context: MHB 
  • Thread starter Thread starter Yankel
  • Start date Start date
  • Tags Tags
    Function Limit
Click For Summary
SUMMARY

The limit of the function \(\frac{x}{\sqrt{x^{2}+2}}\) as \(x\) approaches negative infinity is definitively -1. This conclusion arises from the evaluation of the limit, where the absolute value of \(x\) is considered. As \(x\) approaches -infinity, \(|x| = -x\), leading to the expression simplifying to \(-\frac{1}{\sqrt{1+0}} = -1\). In contrast, the limit as \(x\) approaches positive infinity is 1, as \(|x| = x\) in that scenario.

PREREQUISITES
  • Understanding of limits in calculus
  • Familiarity with the concept of absolute values
  • Knowledge of square roots and their properties
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the properties of limits, particularly one-sided limits
  • Explore the concept of asymptotic behavior in functions
  • Learn about the application of L'Hôpital's rule for indeterminate forms
  • Investigate the implications of absolute values in calculus
USEFUL FOR

Students of calculus, mathematics educators, and anyone seeking to deepen their understanding of limits and function behavior as they approach infinity.

Yankel
Messages
390
Reaction score
0
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 !
 
Physics news on Phys.org
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 $|x|=-x$.

$$(**)=\lim_{x \rightarrow -\infty } \frac{x}{-x\sqrt{1+\frac{2}{x^2} }}=\lim_{x \rightarrow -\infty } \frac{1}{-\sqrt{1+\frac{2}{x^2} }}=-\frac{1}{\sqrt{1+0}}=-1$$
 
Last edited by a moderator:

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 5 ·
Replies
5
Views
1K
Replies
4
Views
1K
  • · Replies 2 ·
Replies
2
Views
10K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K