Question

Guess the value of the limit (correct to six decimal places). (if an answer does not exist, enter dne.) lim hâ0 (4 + h)5 â 1024 h

Answer 1

$\displaystyle\lim_{h\to0}\frac{(4+h)^5-1024}h=\lim_{h\to0}\frac{(4+h)^5-4^5}h$

Recall the definition of the derivative of a function $f(x)$ at a point $x=c$:

$f'(c)=\displaystyle\lim_{h\to 0}\frac{f(c+h)-f(c)}h$

We can then see that $f(c)=c^5$, and by the power rule we have $f'(c)=5c^4$. Then replacing $c=4$, we arrive at

$\displaystyle\lim_{h\to0}\frac{(4+h)^5-4^5}h=5\times4^4=1280$

Alternatively, we could have expanded the binomial, giving

$\dfrac{(4+h)^5-4^5}h=\dfrac{(4^5+5\times4^4h+10\times4^3h^2+10\times4^2h^3+5\times4h^4+h^5)-4^5}h$
$=\dfrac{1280h+640h^2+160h^3+20h^4+h^5}h$
$=1280+640h+160h^2+20h^3+h^4$

and so as $h\to0$ we're left with 1280, as expected.