## anonymous one year ago 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

1. anonymous

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.

2. anonymous

@Hero

3. Hero

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

4. anonymous

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.

5. ytrewqmiswi

Are these 2 equal- 1/(4+x) and (1/4+1/x)

6. ytrewqmiswi

I think u made a mistake in the 1st step

7. Hero

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

8. anonymous

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

9. Hero

You shouldn't use that formula and what you just posted demonstrates why. Instead use this: $$\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}} = \dfrac{a}{b} \div \dfrac{c}{d} = \dfrac{a}{b} \times \dfrac{d}{c}$$

10. anonymous

okay