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  • Need solution for RD Sharma Maths Class 12 Chapter 8 Continuity Excercise 8.1 Question 3

Need solution for RD Sharma Maths Class 12 Chapter 8 Continuity Excercise 8.1 Question 3

Answers (1)

Answer:

                x=3

Hint:

For a function to be continuous at a point, its LHL RHL and value at that point should be equal.

Solution:

Given,

                f(x)=\left(\begin{array}{c} \frac{x^{2}-9}{x-3}, \text { if } x \neq 3 \\\\ 6, \text { if } x=3 \end{array}\right)   

We observe,

[LHL at x=3 ]

\begin{aligned} &\lim _{x \rightarrow 3^{-}} f(x)=\lim _{h \rightarrow 0} f(3-h) \\\\ &\lim _{h \rightarrow 0} \frac{(3-h)^{2}-9}{(3-h)-3}=\lim _{h \rightarrow 0} \frac{3^{2}+h^{2}-6 h-9}{3-h-3} \end{aligned} 

=\lim _{h \rightarrow 0} \frac{h(h-6)}{-h}                                                                    \left[\because(a-b)^{2}=\left(a^{2}+b^{2}-2 a b\right)\right]

\begin{aligned} &=\lim _{h \rightarrow 0}(6-h) \\ &=6 \end{aligned}

[RHL at x=3 ]

\lim _{x \rightarrow 3^{+}} f(x)=\lim _{h \rightarrow 0} f(3+h)

\lim _{h \rightarrow 0} \frac{(3+h)^{2}-9}{(3+h)-3}

=\lim _{h \rightarrow 0} \frac{h^{2}+6 h}{h}                                                                                          \left[\because(a+b)^{2}=\left(a^{2}+b^{2}+2 a b\right)\right]

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{h(h+6)}{h} \\\\ &=\lim _{h \rightarrow 0}(6+h) \\\\ &=6 \end{aligned}

Given

\begin{aligned} &f(3)=6 \\ \\&\therefore \lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3^{-}} f(x)=f(3) \end{aligned}

Hence,f\left ( x \right ) is continuous at x=3.

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