Question

For each of the following find
lim h->0 [f(x+h)-f(x)]/h
f(x)=x^2+1
f(x)=x^2+2x+1
f(x)= x^3

Answer 1

\[\frac{(x+h)^2+1-(x^2+1)}{h}=\frac{x^2+2xh+h^2+1-x^2-1}{h}\]

Answer 2

I'll work the first one for you. The others are up to you.
\[f(x)=x^2+1\]
\[\frac{f(x+h)-f(x)}{h}=\frac{(x+h)^2+1-x^2-1}{h}\]
\[=\frac{x^2+2xh+h^2+1-x^2-1}{h}\]
\[=\frac{2xh+h^2}{h}=\frac{h(2x+h)}{h}=2x+h\]
\[\lim_{h \rightarrow 0}(2x+h)=\lim_{h \rightarrow 0}(2x+0)=2x\]

Answer 3

which equals 2x right for the first one

Answer 4

\[\frac{2xh+h^2}{h}=2x+h\]
As h approaches 0, the limit is 2x

Answer 5

the second one should be h+2x+2 which by the limit would end up being 2x+2

Answer 6

yes, you will learn a quicker method shortly

Answer 7

i am a bit stuck with f(x)=√(x+1)