What is the limit of this function as h approaches 0? $\frac{h}{\sqrt{4 + h} - 2}$ $(h)/(sqrt(4+h)-2)$

Question

is this zero or undefined?

1383 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-02T05:27:19

$L t_{h \to o} \frac{h}{\sqrt{4 + h} - 2}$ $Lt_(h->o)(h)/(sqrt(4+h)-2)$

$= L t_{h \to o} \frac{h (\sqrt{4 + h} + 2)}{(\sqrt{4 + h} - 2) (\sqrt{4 + h} + 2)}$ $=Lt_(h->o)(h(sqrt(4+h)+2))/((sqrt(4+h)-2)(sqrt(4+h)+2)$

$= L t_{h \to o} \frac{h (\sqrt{4 + h} + 2)}{4 + h - 4}$ $=Lt_(h->o)(h(sqrt(4+h)+2))/(4+h-4)$

$=Lt_(h->o)(cancelh(sqrt(4+h)+2))/cancelh " as "h!=0$

$=(sqrt(4+0)+2)=2+2=4$

Answer 2 · 2018-06-06T14:32:45

$4$ .

Recall that, $lim_(h to 0)(f(a+h)-f(a))/h=f'(a)............(ast)$ .

Let, $f(x)=sqrtx," so that, "f'(x)=1/(2sqrtx)$ .

$:. f'(4)=1/(2sqrt4)=1/4$ .

But, $f'(4)=lim_(h to 0)(sqrt(4+h)-sqrt4)/h............[because, (ast)]$ .

$:. lim_(h to 0)(sqrt(4+h)-sqrt4)/h=1/4$ .

$:." The Reqd. Lim."=1/(1/4)=4$ .

Enjoy Maths.!