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  • Need solution for RD Sharma Maths Class 12 Chapter 8 Continuity Excercise 8.1 Question 4

Need solution for RD Sharma Maths Class 12 Chapter 8 Continuity Excercise 8.1 Question 4

Answers (1)

Answer:

                Continuous

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}{cc} \frac{x^{2}-1}{x-1} & , \text { if } x \neq 1 \\\\ 2 & , \text { if } x=1 \end{array}\right)   

We observe,

[LHL at x=1 ]

\begin{aligned} &\lim _{h \rightarrow 0} \frac{(1-h)^{2}-1}{(1-h)-1} \\\\ &=\lim _{h \rightarrow 0} \frac{1+h^{2}-2 h-1}{1-h-1} \end{aligned}

=\lim _{h \rightarrow 0} \frac{h^{2}-2 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-2)}{-h} \\\\ &=\lim _{h \rightarrow 0}(2-h) \\\\ &=2 \end{aligned}

[RHL at x=1]

\begin{aligned} &\lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h) \\\\ &\lim _{h \rightarrow 0} \frac{(1+h)^{2}-1}{(1+h)-1} \end{aligned}

=\lim _{h \rightarrow 0} \frac{h^{2}+2 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+2)}{h} \\\\ &=\lim _{h \rightarrow 0}(2+h) \\\\ &=2 \end{aligned}

Given

\begin{aligned} &f(1)=2 \\\\ &\therefore \lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{-}} f(x)=f(1) \end{aligned}

Hence, f\left ( x \right ) is continuous at x=1.

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