Question

evaluate the limit lim x->0 (5+x)^-1-5^-1/(x). The answer is -1/25 but I don't know how.

Answer 1

\[\lim_{x \rightarrow 0}((5+x)^{-1}-5^{\frac{-1}{x}})\]
correct?

Answer 2

" The answer is 1/25" <--- seems that's found already =)

Answer 3

the answer is 1/25 but I don't
know how

Answer 4

I don't think I'm seeing the question correctly, could be wrong?
But I don't the above I think I wrote has a limit of 1/25.

Answer 5

\[\lim_{x \rightarrow 0}\frac{(5+x)^{-1}-5^{-1}}{x}\]?

Answer 6

yes

Answer 7

do you know derivatives
because you can skip a lot of steps if you do

Answer 8

I do know them

Answer 9

\[\lim_{x \rightarrow a}\frac{f(x)-f(a)}{x-a}=f'(x)|_{x=a}\]
are you familiar with this definition?

Answer 10

sort of, but I'm more familiar with f(x+h)-f(x)/h

Answer 11

\[\lim_{x \rightarrow 0}\frac{(5+x)^{-1}-(5+0)^{-1}}{x-0}=\lim_{x \rightarrow a}\frac{f(x)-f(a)}{x-a}\]
If you are good at comparing you will be able to see f(x)=(5+x)^(-1) and a=0

Answer 12

\[\lim_{x \rightarrow 0}\frac{(5+x)^{-1}-(5+0)^{-1}}{x-0}=[(5+x)^{-1}]'|_{x=0}\]

Answer 13

so you will have to know chain rule to do it this way.

Answer 14

If you don't like this way we can do it the good old fashion way.

Answer 15

yes please, if its not any trouble

Answer 16

and the answer is -1/25, sorry

Answer 17

\[\lim_{x \rightarrow 0}\frac{\frac{1}{5+x}-\frac{1}{5}}{x}=\lim_{x \rightarrow 0}\frac{1}{x}(\frac{1}{x+5}-\frac{1}{5})\]
We cannot just plug in 0 because the bottom will be 0.
We want to get rid of that factor of x on the bottom.
So sometimes a trick when seeing separate fractions is to combine them.

Answer 18

An easy way to combine fractions is: