Find the Limit

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  • Thread starter nycmathdad
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  • #1
nycmathdad
74
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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?
 

Answers and Replies

  • #2
jonah1
108
0
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

Do try to take the time to read your book instead of insisting on your ineffective method:
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?
 
Last edited:
  • #3
HOI
923
2
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.
 
  • #4
nycmathdad
74
0
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.
 
  • #5
HOI
923
2
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.
 
  • #6
nycmathdad
74
0
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.

Ok. I will practice more problems before moving on in the textbook.
 

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