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

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

Answers (1)

Answer:

                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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