3 years ago

HELP???

Mathematics

1 answer:

zzz [600]3 years ago

6 0

D(t)=25(0.82)^t

d(4)=11.3mg

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

34kurt

$\displaystyle\lim_{h\to0}\frac{(5+h)^{-1}-5^{-1}}h=\lim_{h\to0}\frac{\frac5{5(5+h)}-\frac{5+h}{5(5+h)}}h$
$\displaystyle=-\lim_{h\to0}\frac h{5(5+h)h}$
$\displaystyle=-\lim_{h\to0}\frac1{5(5+h)}=-\frac1{25}$

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.

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