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svetoff [14.1K]
3 years ago
15

Evaluate the limit, if it exists. (if an answer does not exist, enter dne.) lim h→0 (5 + h)−1 − 5−1 h

Mathematics
1 answer:
34kurt3 years ago
4 0
\displaystyle\lim_{h\to0}\frac{(5+h)^{-1}-5^{-1}}h=\lim_{h\to0}\frac{\frac5{5(5+h)}-\frac{5+h}{5(5+h)}}h
\displaystyle=-\lim_{h\to0}\frac h{5(5+h)h}
\displaystyle=-\lim_{h\to0}\frac1{5(5+h)}=-\frac1{25}

Alternatively, recall that if f(x)=\dfrac1x, then f'(x)=-\dfrac1{x^2}, and so

f'(5)=\displaystyle\lim_{x\to5}\frac{\frac1x-\frac15}{x-5}

Take h=x-5, so that x=h+5, and we have the original limit. So the limit is equivalent to the value of f'(5), i.e.

f'(5)=-\dfrac1{5^2}=-\dfrac1{25}
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