Question

Find the limit Picture below

Answer 1

The limit doesn’t exist

Answer 2

Answer:

a. 4

Step-by-step explanation:

We want to find;

$\lim_{h \to 0} \frac{f(2+h)-f(2)}{h}$

If $f(x)=x^2$.

$f(2+h)=(2+h)^2$

$f(2+h)=4+4h+h^2$

Also;

$f(2)=2^2$

$f(2)=4$

This implies that;

$\lim_{h \to 0} \frac{4+4h+h^2-4}{h}$

Simplify;

$\lim_{h \to 0} \frac{4h+h^2}{h}$

Factor to get;

$\lim_{h \to 0} \frac{h(4+h)}{h}$

Cancel the common factors.

$\lim_{h \to 0} 4+h$

Substitute h=0 to get

$\lim_{h \to 0} 4+h=4+0=4$