Limit of a function when x goes to -infinity

In summary: $$\lim_{x \rightarrow -\infty } \frac{x}{\sqrt{x^2+2}}=\lim_{x \rightarrow -\infty } \frac{1}{\sqrt{1+\frac{2}{x^2} }}=-\frac{1}{\sqrt{1+0}}=-1$$ $$\lim_{x \rightarrow -\infty } \frac{x}{\sqrt{x^2+2}}=\lim_{x \rightarrow -\infty } \frac{-1}{\sqrt{1+\frac{2}{x^2} }}
  • #1
Yankel
395
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
  • #2
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:
  • #3
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:

FAQ: Limit of a function when x goes to -infinity

What does it mean for a function to have a limit as x approaches negative infinity?

When we say a function has a limit as x approaches negative infinity, it means that as x gets smaller and smaller (closer to negative infinity), the output values of the function approach a certain value. This value is called the limit and it represents the behavior of the function as x approaches negative infinity.

How do you determine the limit of a function when x goes to -infinity?

To determine the limit of a function as x approaches negative infinity, we plug in increasingly small negative values for x and observe the corresponding output values. If the output values approach a certain value, that value is the limit of the function. If the output values do not approach a single value, the limit does not exist.

Can a function have a limit as x approaches negative infinity if it is undefined at x = -infinity?

No, a function cannot have a limit as x approaches negative infinity if it is undefined at x = -infinity. In order for a function to have a limit, it must approach a single value as x gets closer and closer to the given point.

Can a function approach different values from the left and right as x approaches negative infinity?

Yes, it is possible for a function to approach different values from the left and right as x approaches negative infinity. This is known as a one-sided limit and it can occur if the function has different behaviors on either side of the given point.

How can the limit of a function at x = -infinity be used to determine the end behavior of the function?

The limit of a function at x = -infinity can be used to determine the end behavior of the function. If the limit is a finite number, this indicates that the function approaches a horizontal asymptote as x goes to negative infinity. If the limit is positive or negative infinity, this indicates that the function increases or decreases without bound as x approaches negative infinity, respectively.

Similar threads

I A question about limits and infinity
Replies
7
Views
2K
MHB Limit of Trigonometric Function....2
Replies
1
Views
964
MHB How to Find the Limit of (1 - x)/[(3 - x)^2] as x Approaches 3?
Replies
5
Views
994
MHB What is the Limit of (5x)/(100 - x) as x Approaches 100 from the Left?
Replies
4
Views
1K
MHB Limit of = (sin nx) / (sin x) as n goes to infinity.
Replies
2
Views
9K
MHB Derivative and Limit of an Exponential Function
Replies
3
Views
2K
B Question about the limits of integration in a change of variables
Replies
2
Views
1K
MHB Limit involving a hyperbolic function
Replies
4
Views
1K
MHB Is the Limit of 1/(x^2 - 9) as x Approaches -3 from the Left Positive Infinity?
Replies
5
Views
1K
MHB Find the limit as x goes to infinity
Replies
5
Views
2K

Hot Threads

Recent Insights

Back
Top