Question

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

The answer is -1/16

Here is what I did: (1/4 + 1/x)(1/4 + 1/x)
1/16+2/4x+1/x^2
1/16+2/4(-4)+1/(-4)^2
1/16-2/16+1/16
0

Answer 1

I know this guy solved it:
https://www.youtube.com/watch?v=9_RUnj-5wEk,
but I don't get why my technique doesn't work.

Answer 2

@Hero

Answer 3

\(\begin{align*}\dfrac{\dfrac{1}{4} + \dfrac{1}{x}}{4 + x} &= \left(\dfrac{1}{4} + \dfrac{1}{x}\right) \div (4 + x) \\&= \left(\dfrac{x}{4x} + \dfrac{4}{4x}\right) \div (4 + x) \\&= \left(\dfrac{x + 4}{4x} \times \dfrac{1}{x + 4}\right)\\&=\dfrac{1}{4x}\end{align*}\)

\(\lim_{x \to -4}\dfrac{1}{4x} = \dfrac{1}{4(-4)} = -\dfrac{1}{16}\)

Answer 4

Thank you! But, I was more wondering why my answer didn't work... because, I know there is something wrong about it, but I can't spot what.

Answer 5

Are these 2 equal-

1/(4+x) and (1/4+1/x)

Answer 6

I think u made a mistake in the 1st step

Answer 7

@vvbb you can discover what you did wrong for yourself.

Answer 8

Can't I use this formula for the first step? (a/b)/(c/d)=(a/b)*(c/d)

Answer 9

You shouldn't use that formula and what you just posted demonstrates why.

Instead use this:

\(\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}} = \dfrac{a}{b} \div \dfrac{c}{d} = \dfrac{a}{b} \times \dfrac{d}{c}\)

Answer 10

okay