Question

find the limit if it exists :lim x->0 (sqrt(1+x)-1)/x. We must multiply by the conjugate first and cannot use L'hospitals rule.

Answer 1

So here is our conjugate multiplication,
\[\large\rm \lim_{x\to0}\frac{\sqrt{1+x}-1}{x}\color{royalblue}{\left(\frac{\sqrt{1+x}+1}{\sqrt{1+x}+1}\right)}\]

Answer 2

Remember what you get when you multiply conjugates?

Answer 3

Leave the bottom alone for now,\[\Large\rm \lim_{x\to0}\frac{\left(\sqrt{1+x}-1\right)\left(\sqrt{1+x}+1\right)}{x\left(\sqrt{1+x}+1\right)}\]

Answer 4

What do we get in the numerator?

Answer 5

1+x-1

Answer 6

\[\Large\rm \lim_{x\to0}\frac{1+x-1}{x\left(\sqrt{1+x}+1\right)}\]Ok good :o
Know how to proceed?

Answer 7

kind of, the 1's cancel on top, and the x cancels with the denoms x(?) then we are left with 1/sqrt1+sqrtx+1. replace the x with 0 and that leaves us with 1/sqrt1+1

Answer 8

Yes great!
So canceling out the 1's and x's allowed us to plug in x=0 without any problems.

Answer 9

cool, Thanks Zepdrix!