Question

lim h→0 [1/ (h + 5)^2 − 1/ 25]/ h

Answer 1

\[\lim_{h\to0}\frac{\dfrac{1}{(h+5)^2}-\dfrac{1}{25}}{h}=\lim_{h\to0}\frac{\dfrac{25}{25(h+5)^2}-\dfrac{(h+5)^2}{25(h+5)^2}}{h}=\lim_{h\to0}\frac{\dfrac{25-(h+5)^2}{25(h+5)^2}}{h}\]
Try expanding the numerator's numerator, you should get some reducible terms.

Answer 2

\[\mathbb{\color{blue}{Welcome~To~Openstudy}}\]
It looks like Siths has given you a great explanation, what else do you need?

Answer 3

I just keep getting 0 when i reduce the terms... I'll try again

Answer 4

\[\lim_{h\to0}\frac{\dfrac{25-(h+5)^2}{25(h+5)^2}}{h}=\lim_{h\to0}\frac{\dfrac{25-(h^2+10h+25)}{25(h+5)^2}}{h}=\lim_{h\to0}\frac{\dfrac{-h^2-10h}{25(h+5)^2}}{h}\]
which further reduces to
\[\lim_{h\to0}\frac{-h-10}{25(h+5)^2}\]