Question

lim x tends to 0 (4+x)²-16/x​

Answer 1

Step-by-step explanation:

We have,

$\lim_{x \rarr0} \frac{(4 + x) ^{2} - 16 }{x}$

$= \lim_{x \rarr0} \frac{16 + 8x + {x}^{2} - 16}{x}$

$= \lim_{x \rarr0} \frac{x(x + 8)}{x}$

$= \lim_{x \rarr0}(x + 8)$

$= 8$