# Answer the following questions: Please refer to the questions in the image: Thank you?

May 23, 2018

a) $\textcolor{b l u e}{{\lim}_{x \rightarrow {3}^{-}} \frac{x + 3}{{x}^{2} - 9} = - \infty}$
b)color(red)[lim_(xrarr-4^+)(9x)/(16-x^2)=oo

#### Explanation:

i will solve question 1

a) $\textcolor{b l u e}{{\lim}_{x \rightarrow {3}^{-}} \frac{x + 3}{{x}^{2} - 9}}$

take a number from left of 3 like 2.9

now offset it in the limit ${\lim}_{x \rightarrow {3}^{-}} \frac{x + 3}{{x}^{2} - 9}$

if the sign of the limit equal $-$ that mean the value of limit equal $- \infty$ ,but if the sign of the limit equal $+$ that mean the limit equal $+ \infty$

${\lim}_{x \rightarrow {3}^{-}} \frac{x + 3}{{x}^{2} - 9} = \frac{2.9 + 3}{8.41 - 9} = \frac{+}{-} = -$

${\lim}_{x \rightarrow {3}^{-}} \frac{x + 3}{{x}^{2} - 9} = - \infty$

b)color(red)[lim_(xrarr-4^+)(9x)/(16-x^2)

take a number from right of -4 like -4.1

offset it in the limit ${\lim}_{x \rightarrow - {4}^{+}} \frac{9 x}{16 - {x}^{2}}$

we will get

${\lim}_{x \rightarrow - {4}^{+}} \frac{9 x}{16 - {x}^{2}} = \frac{- 36.9}{16 - 16.81} = \frac{-}{-} = +$

${\lim}_{x \rightarrow - {4}^{+}} \frac{9 x}{16 - {x}^{2}} = \infty$