Question

please explain for medal! (:

Answer 1

:)

Answer 2

\[\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}=f'(x)\]

Answer 3

or you could just evaluate the limit

Answer 4

\[\lim_{h \rightarrow 0}\frac{(5(x+h)^2-2)-(5x^2-2)}{h}\]

Answer 5

try to simplify the stop you can start by looking at the -2+2 part

Answer 6

top * not the stop

Answer 7

so the denominator would be 0 so (5x^2-2)-(5x^2-2)

Answer 8

we can't plug in 0 yet

Answer 9

because we would get 0/0

Answer 10

oh oops

Answer 11

I'm asking you to try to simplify the top
the starting point is looking at -2+2

Answer 12

then multiplying the (x+h)^2 out

Answer 13

then you will some other things zero out on top

Answer 14

okk one sec

Answer 15

\[5(x+h)^2-2-(5x^2-2) \\ =5(x+h)^2-2-5x^2+2 \\ 5(x+h)^2-5x^2\]
try that again

Answer 16

(x+h)(x+h) is?

Answer 17

x^2+xh+xh+h^2

Answer 18

\[5(x+h)^2-2-(5x^2-2) \\ =5(x+h)^2-2-5x^2+2 \\ =5(x+h)^2-5x^2\\ =5(x^2+2xh+h^2)-5x^2 \]
okay distribute the 5 and combine any like terms

Answer 19

10xh+5h^2

Answer 20

the 5x^2 cancels

Answer 21

\[\lim_{h \rightarrow 0}\frac{(5(x+h)^2-2)-(5x^2-2)}{h}=\lim_{h \rightarrow 0}\frac{10xh+5h^2}{h}\]
do you think you see where to go from here?

Answer 22

because you cancel the h(10x+5h)/h=10x+5h
and yea that is 10x since the h->0

Answer 23

thank you!