#### Please solve RD Sharma class 12 chapter 8 Continuity exercise Fill in the blanks question 1 maths textbook solution

Answer: $3a^{2}$

Hint:

Use the formula $(x^{3}-a^{3})$ so that $f(x)\neq \frac{0}{0}$ at $x\neq a$

Given:

$f(x)=\left\{\begin{array}{cl} \frac{x^{3}-a^{3}}{x-a} & , x \neq a \\ b & , x=a \end{array}\right.$   is continuous at $x=a$

Solution:

If $f(x)$ is continuous at $x=a$ , then RHL= LHL

LHL,

\begin{aligned} &\lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow a} \frac{x^{3}-a^{3}}{x-a} \\ &=\lim _{x \rightarrow a^{-}} \frac{(x-a)\left(x^{2}+a x+a^{2}\right)}{(x-a)} \\ &=\lim _{x \rightarrow a^{-}} x^{2}+a x+a^{2} \\ &=\lim _{\mathbb{\Xi} \rightarrow 0} a^{2}+a^{2}+a^{2} \\ &=3 a^{2} \end{aligned}$\left [ \because x=a \right ]$

RHL,

\begin{aligned} &\lim _{x \rightarrow a^{+}} f(x)=\lim _{x \rightarrow a^{+}} b \\ &=b \end{aligned}

As $f(x)$ is continuous at $x=a$, RHL = LHL

$b=3a^{2}$
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 14, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}