Question

How do i do this without L'hopitals rule?

Answer 1

lim h → 0 [sqrt(16 + h) − 4 ]/h

Answer 2

Multiply the top and bottom by the conjugate?

Answer 3

multiply by the conjugate to get rid of the sqrt

Answer 4

recognize it as the derivative of \(f(x)=\sqrt{x}\) at \(x=16\)

Answer 5

Remember the conjugate is \(\sqrt{16+h}+4\)

Answer 6

if you know that the derivative of \(\sqrt{x}\) is \(\frac{1}{2\sqrt{x}}\) then replace \(x\) by \(16\) because this is the derivative

Answer 7

If you multiply by conjugate though, you end up with \[
\lim_{h\to 0}\frac{(16+h)-4^2}{h(\sqrt{16+h}+4)}
\]

Answer 8

4^2 is 16 so it cancels out and then your h's will cancel as well

Answer 9

It simplifies to \[
\lim_{h\to 0}\frac{1}{\sqrt{16+h}+4}
\]Which is continuous at \(h=0\)