Question about a limit problem, Calculus I

herbpixie

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This is my problem:

lim x--> -2 ((1/(x+2))+(4/((x^2)-4))

I know that the answer is 1/4, but I keep getting 4. This is my work so far:

((x^2)+4)/((x+2)((x^2)-4)) + (4(x+2)/((x+2)((x^2)-4))

I factor (x^2)-4 into (x+2)(x-2) in every place it appears. That causes everything in the denominator to cancel out, leaving 4 in the numerator.

I am obviously doing something wrong, but I've tried 5 different ways to solve this problem, and I end up with 4 every time. Can someone please tell me what I am doing wrong? I would be extremely grateful.
 
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Why don't you try a LEAST common denominator?

The left term

\(\displaystyle \frac{1}{x+2}\cdot \frac{x-2}{x-2}\)

The answer is NOT 1/4. The answer is also not 4.

On the other hand, you may be working on the wrong problem. Try x ==> -2 instead.
 
Bugger, it's a typo. It was supposed to be lim x--> -2

Why isn't the least common denominator (x+2)((x^2)-4)? At that point, I want to expand ((x^2)-4) to (x+2)(x-2), which cancels out everything in the numerator except for the 4.

Thank you so much for your help. I took precal two semesters ago after not taking any math for 20 years. I ended up with a B, but I still have trouble with some things that I should probably already know. I really appreciate it.
 
x+2 is a factor of x^2 - 4.

What is the least common multiple of 6 and 12? Hint: It's NOT 72.

Observation: It doesn't HAVE to be the LEAST common multiple. It should work with any common multiple. The LEAST just simplifies your life.
 
Even if I use (x+2)(x-2) as the denominator, I end up with the wrong answer. Here is my bad math. :)

lim x--> -2 (1/(x+2))+(4/((x^2)-4))

= (1/(x+2))+(4/((x+2)(x-2))

= ((x+2)+4)/((x+2)(x-2)) I have a feeling that my denominator is still incorrect?

= 4/(x-2) since the (x+2) cancels out

This leaves me with the limit as -1, not 1/4.

Again, thank you. You rock.
 
This is my problem:
lim x-->2 ((1/(x+2))+(4/((x^2)-4))
I know that the answer is 1/4, but I keep getting 4.
As written
\(\displaystyle \displaystyle\lim_{x\to 2}\frac{1}{x+2}+\frac{4}{x^2-4}\) has no limit.
Please check the question.
 
As written
\(\displaystyle \displaystyle\lim_{x\to 2}\frac{1}{x+2}+\frac{4}{x^2-4}\) has no limit.
Please check the question.

Hi pka,

There was a typo in the original question that I corrected in the comments. The question should have been:

lim --> -2 (1/(x+2))+(4/((X^2)-4))

It was correct, except that the limit should be approaching -2, not 2. My apologies for the confusion. I have edited the original post.
 
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There was a typo in the original question that I corrected in the comments. The question should have been:
lim --> -2 (1/(x+2))+(4/((X^2)-4)).
Please try to post correctly.
\(\displaystyle \dfrac{1}{x+2}+\dfrac{4}{(x^2-4)}=\dfrac{1}{x+2}+\dfrac{4}{(x+2)(x-2)}\)
\(\displaystyle = \dfrac{x+2}{(x+2)(x-2)}= \dfrac{1}{(x-2)}\)

Now the limit is \(\displaystyle -\dfrac{1}{4}.\)
 
Please try to post correctly.
\(\displaystyle \dfrac{1}{x+2}+\dfrac{4}{(x^2-4)}=\dfrac{1}{x+2}+\dfrac{4}{(x+2)(x-2)}\)
\(\displaystyle = \dfrac{x+2}{(x+2)(x-2)}= \dfrac{1}{(x-2)}\)

Now the limit is \(\displaystyle -\dfrac{1}{4}.\)

I can assure you that the mistake was unintentional. I'm having trouble doing this in text, and I don't know how to do the html code that shows the actual fractions.

I don't understand what happens to the 4 in the numerator. I'm copying your code in hopes that it works. I'm with you at this point:

\(\displaystyle \dfrac{1}{x+2}+\dfrac{4}{(x^2-4)}=\dfrac{1}{x+2}+\dfrac{4}{(x+2)(x-2)}\)

I understand how you got the denominator in the next step, but I don't understand why the numerator doesn't become (x-2) + 4 when the two fractions are added. Since the denominator in the second fraction is already correct, why wouldn't you multiply 1 * (x-2) to get the numerator in the first fraction? Thank you for your help.
 
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I can assure you that the mistake was unintentional.
I made no mistake
The limit is \(\displaystyle -\dfrac{1}{4}\).
Your textbook/instructor may be mistaken, but I am not.
Look at this web page.
Y
ou should learn to use that web resource!
 
I made no mistake
The limit is \(\displaystyle -\dfrac{1}{4}\).
Your textbook/instructor may be mistaken, but I am not.
Look at this web page.
Y
ou should learn to use that web resource!

I meant that my mistake was unintentional. I did not mean to imply that you made one. I'm just trying to understand how to work the problem, and I'm not understanding how you got the numerator. The website doesn't take me step by step through the problem. While the graph is helpful, it doesn't answer my question.
 
Yes is does if you click "show steps'.

I did, but it still skips the step that I'm confused about. When I click "show steps," it just shows me the original equation in a different order. I don't understand why the equation does not become this when you find the common denominator:

((x+2)/(x+2)(x-2))+(4/((x+2)(x-2)))

= 4/(x-2)
 
I did, but it still skips the step that I'm confused about. When I click "show steps," it just shows me the original equation in a different order. I don't understand why the equation does not become this when you find the common denominator:
((x+2)/(x+2)(x-2))+(4/((x+2)(x-2)))= 4/(x-2)
Look, we here are not to be held responsible for your lack of basic knowledge. These are basic middle school mathematics issues. If you lack those skills, then that is not our problem.
 
Look, we here are not to be held responsible for your lack of basic knowledge. These are basic middle school mathematics issues. If you lack those skills, then that is not our problem.

Thanks anyway. Courtesy is a lost art, and you could be a lot nicer. Perhaps if you'd read my other comments, you'd see why there are holes in my knowledge. I won't be coming to this site again. I can get trolled on youtube.
 
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Thank you, JeffM. I really appreciate it. I finally figured it out last night watching patrickjmt videos. I was trying to cancel out before finishing adding on the top.
 
I can assure you that the mistake was unintentional. I'm having trouble doing this in text, and I don't know how to do the html code that shows the actual fractions.

I don't understand what happens to the 4 in the numerator. I'm copying your code in hopes that it works. I'm with you at this point:

\(\displaystyle \dfrac{1}{x+2}+\dfrac{4}{(x^2-4)}=\dfrac{1}{x+2}+\dfrac{4}{(x+2)(x-2)}\)

I understand how you got the denominator in the next step, but I don't understand why the numerator doesn't become (x-2) + 4 when the two fractions are added. Since the denominator in the second fraction is already correct, why wouldn't you multiply 1 * (x-2) to get the numerator in the first fraction? Thank you for your help.

The numerator DOES become (x - 2) + 4.

Now, SIMPLIFY that....

(x - 2) + 4 = x -2 + 4 = x + 2 ........combine the -2 and the + 4.

I hope that helps.
 
The numerator DOES become (x - 2) + 4.

Now, SIMPLIFY that....

(x - 2) + 4 = x -2 + 4 = x + 2 ........combine the -2 and the + 4.

I hope that helps.

Thank you--I figured it out, but that was exactly what I was looking for. I'm in my second semester of math after a 20 year break, so I'm rusty on some simple things. I appreciate your help.
 
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