Question

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Answer 1

Is this your question?
\[\lim_{x\to 0} \frac{1}{x+4}- \frac{1}{4x} \]

Answer 2

\[\lim_{x \rightarrow 0}\frac{4x-(x+4)}{4x(x+4)}=\lim_{x \rightarrow 0}\frac{3x-4}{4x(x+4)}\]
We see nothing cancels and we plug 0 in we get -4/0 which means the limit does not exist (there was no way to manipulate the function making it continuous at x=0)

Answer 3

what do you mean how did he get that?

Answer 4

the pretty symbols?

Answer 5

No, I want to see the procedure. I kind of forgot how to solve the fraction. Since the limit does not exist, how about this problem: lim x->0 [1/(2+x)]-(1/2)/x

Answer 6

Combine the fractions.

Answer 7

and see if anything cancels

Answer 8

\[\frac{1}{2+x}-\frac{1}{2x}=\frac{2x-(2+x)}{2x(2+x)}=\frac{x-2}{2x(x+2)}\]

Answer 9

Okay, I just solved the last problem. The answer is -1/4, correct?

Answer 10

same thing as x->0 the limit does not exist

Answer 11

Are you sure you are typing these problems in correctly?

Answer 12

Yes, straight from the textbook.

Answer 13

so what i have is exactly what it says?

Answer 14

Here you go.

Answer 15

ok so those are the wrong problems above

Answer 16

I am sorry for the confusion. :(

Answer 17

\[\lim_{x \rightarrow 0}\frac{\frac{1}{x+4}-\frac{1}{4}}{x}\]
If I were writing this without pretty latex like I just did I would say:
lim x->0 [[1/(x+4)]-(1/4)]/x
so I can know where the numerator is and the denominator

Answer 18

\[\frac{a}{b}+\frac{c}{d}=\frac{a d+cb}{bd}\] recall this

Answer 19

\[\lim_{x \rightarrow 0}\frac{\frac{4-(x+4)}{4(x+4)}}{x}\]

Answer 20

\[\lim_{x \rightarrow 0}\frac{-x+4-4}{4(x+4)} \cdot \frac{1}{x}=\lim_{x \rightarrow 0}\frac{-x}{4(x+4)} \cdot \frac{1}{x}\]

Answer 21

\[\lim_{x \rightarrow 0}\frac{-1}{4(x+4)}\]

Answer 22

now guess what you get to do now?

Answer 23

I got -1/16. Is that right?

Answer 24

yes! :)

Answer 25

So I didn't forget how to solve the fraction then. Thank you so much!

Answer 26

lol

Answer 27

great! :)