## anonymous 4 years ago lim h → 0 (f(2+h)-f(2))/h F(x)=x^2+1

1. anonymous

4 and in a week you will do this problem in your head in two seconds by saying "the derivative of $x^2+1$ is $2x$ and $2\times 2=4$ but you have not gotten there yet so you have some work to do

2. anonymous

Could you please show work on how to solve this using the equation, so I know for future refrences?

3. anonymous

$f(2)=2^2+1=5$ $f(2+h)=(2+h)^2+1=4+4h+h^2+1=5+4h+h^2$ $f(2+h)-f(2)=5+4h+h^2-5=4h+h^2$ $\frac{f(2+h)-f(2)}{h}=\frac{4h+h^2}{h}=\frac{h(4+h)}{h}=4+h$

4. anonymous

now you can take the limit as h goes to zero and you get 4 for sure

5. anonymous

all steps are there, nothing left out

6. anonymous

WOW thank you!

7. anonymous

yw notice that the 5's add to zero. it will always work this way if the limit exists. because you have an h in the denominator, the only way for this limit to exist is if the constant goes and you have nothing but h's