Question

Which of the following function can be defined at indicated point so that the resulting function is continuous -

• A. $f\left(x\right)=\frac{x^2-2x-8}{x+2}$ at $x=-2$
• B. All of these
• C. $f\left(x\right)=\frac{x^3+64}{x+4}$ at $x=-4$
• D. $f\left(x\right)=\frac{3-\sqrt{x}}{9-x}$ at $x=9$

Answer 1

The correct option is B All of these

for $f^n$ defined its limit showed exist at that pt

(A) $\underset{x\to -2}{lim}\frac{x^2-2x-8}{x+2}=\frac{0}{0}\Rightarrow \underset{x\to -2}{lim}\frac{2x-2}{1}=-6$

(B) $f\left(x\right)\{\begin{array}{c}1x>7\\ -1x<7\end{array}$ so $lim$ do not exists

(C) $\underset{x\to -4}{lim}\frac{x^3+64}{x+4}=\%3x^2\Rightarrow 3(-4{)}^2=3(16)=48$ finite

(D) $\underset{x\to 9}{lim}\frac{3-\sqrt{x}}{9-x}=\frac{-\frac{1}{2\sqrt{x}}}{-1}=\frac{1}{2\sqrt{9}}=\frac{1}{6}$

Ans will be $A,C,D$