Need solution for RD Sharma Maths Class 12 Chapter 18 Continuity Excercise 8.1 Question 36 Subquestion (vi)

$k=10$

Hint:

$f\left ( x \right )$  must be defined. The limit of the $f\left ( x \right )$ approaches the value $x$ must exist.

Given:

$f(x)=\left(\begin{array}{cc} \frac{x^{2}-25}{x-5} & , x \neq 5 \\\\ k & , x=5 \end{array}\right)$

Solution:

$f(x)=\left(\begin{array}{cc} \frac{x^{2}-25}{x-5} & , x \neq 5 \\\\ k & , x=5 \end{array}\right)$

$f(x)=\left(\begin{array}{cc} \frac{(x-5)(x+5)}{x-5} \\\\ k, x=5 \end{array}, x \neq 5\right)$                                  $\left[\because a^{2}-b^{2}=(a+b)(a-b)\right]$

$f(x)=\left(\begin{array}{cc} (x+5), & x \neq 5 \\\\ k & , x=5 \end{array}\right)$

If $f\left ( x \right )$ is continuous at $x=5$ , then

\begin{aligned} &\lim _{x \rightarrow 5} f(x)=f(5) \\ &\lim _{x \rightarrow 5}(x+5)=k \\ &k=5+5=10 \\ &k=10 \end{aligned}

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