stuck on this limit problem

[PaulKraemer]

Hi,

I am stuck on the following limit problem:

lim as h -> 0 of
( (2+h)^(-2) - (2)^(-2) ) / h

I assume I have to multiply the numerator and denominator by the same thing to try to eliminate the h in the denominator, but I can't figure out how to get started on this problem. Any help or hints would be greatly appreciated.

Thanks in advance,
Paul

 

[soroban]

Hello, Paul!

No, we don't need a "conjugate" for this one . . . just old-fashioned algebra.

\(\displaystyle \lim_{h\to0}\frac{(2+h)^{-2} - (2)^{-2}}{h}\)

\(\displaystyle \text{We have: }\;\frac{\dfrac{1}{(2+h)^2} - \dfrac{1}{2^2}}{h} \;=\;\frac{\dfrac{2^2-(2+h)^2} {2^2(2+h)^2}}{h} \;=\; \frac{4 - (4 + 4h + h^2)}{4h(2+h)^2} \;=\;\frac{4 - 4 - 4h - h^2}{4h(2+h)^2}\)

. . . . . . \(\displaystyle =\;\frac{-4h - h^2}{4h(2+h)^2} \;=\;\frac{-h(4+h)}{4h(2+h)^2} \;=\; \frac{-(4+h)}{4(2+h)^2}\)


\(\displaystyle \text{Therefore: }\:\lim_{h\to0} \frac{-(4+h)}{4(2+h)^2} \;=\; \frac{-(4+0)}{4(2+0)^2} \;=\;\frac{-4}{4(4)} \;=\;-\frac{1}{4}\)

 

[PaulKraemer]

Thank you Soroban!

That was a big help!

Paul