Find the Limit

• MHB
Find the limit of x/(x^2 - 4) as x tends to 2 from the right.

If I plug x = 0, I will get 0/-4 = asymptote. Again, is graphing the best to do this one?

I can also create a number line.

<----------(-2)----------(0)---------------(2)-------->

I can then select values for x from each interval. I then plug those values into the given function to see if the intervals are positive or negative.

True?

jonah1
Beer soaked fill in the boxes hint follows.
Find the limit of x/(x^2 - 4) as x tends to 2 from the right.

If I plug x = 0, I will get 0/-4 = asymptote. Again, is graphing the best to do this one?

I can also create a number line.

<----------(-2)----------(0)---------------(2)-------->

I can then select values for x from each interval. I then plug those values into the given function to see if the intervals are positive or negative.

True?
As x approaches 2 from the $\boxed{?}$, x approaches $\boxed{?}$ and
$x^2−4$ approaches $\boxed{?}$ from the $\boxed{?}$. Therefore,
the ratio $\frac{x}{x^2−4}$ becomes $\boxed{?}$ in the $\boxed{?}$ $\boxed{?}$, so $$\mathop {\lim }\limits_{x\to2^+} \frac{x}{{x^2-4}} = \boxed{?}$$

Note: Number of question marks corresponds to number of letters or symbols.

Again, to support whatever conclusion you get from filling in the boxes, make a table with the following values of x: 2.1, 2.01, 2.001

2. I don't have time to read the textbook lessons. I usually make use of the chapter outline as my guide. For example, Section 1.5 is all about Limits at Infinity. I then search You Tube for Limits at Infinity video lessons. I take notes on everything said in the video lesson. I work out all sample questions with the video instructor. Is this a good way to learn the material?

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HOI
For x> 2 $x^2- 4$ is positive. As x approaches 2 from the right, $\frac{x}{x^2- 4}$ goes to $+\infty$.
For x< 2 $x^2- 4$ is negative. As x approaches 2 from the -eft, $\frac{x}{x^2- 4}$ goes to $+\infty$.

x= 2 is a vertical asymptote.

For x> 2 $x^2- 4$ is positive. As x approaches 2 from the right, $\frac{x}{x^2- 4}$ goes to $+\infty$.
For x< 2 $x^2- 4$ is negative. As x approaches 2 from the -eft, $\frac{x}{x^2- 4}$ goes to $+\infty$.

x= 2 is a vertical asymptote.

So the answer is positive infinity.

HOI
No. I miswrote (I'm doing that too often lately!) . I apologize for that.

I meant to say "For x< 2 $x^2- 4$ is negative, As a approaches 2 from the left, $\frac{x}{x^2- 4}$ goes to $- \infty$". NEGATIVE infinity, not positive infinity. $\frac{x}{x^2- 4}$ does NOT have a limit as x goes to 2.

I meant to say "For x< 2 $x^2- 4$ is negative, As a approaches 2 from the left, $\frac{x}{x^2- 4}$ goes to $- \infty$". NEGATIVE infinity, not positive infinity. $\frac{x}{x^2- 4}$ does NOT have a limit as x goes to 2.