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NDA Math Limit, Continuity and Differentiability

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Consider the function

f(x)=\begin{cases} x^2-5, & x\leq 3 \\ \sqrt{x+13}, & x>3 \end{cases}

[NDA II 2014]

Question : 88 of 105

Marks: +1, -0

\text{ What is }\;\lim\limits_{x \rightarrow 3}f(x)\;\text{ equal to }?

Solution:

f(x)=\begin{cases} x^{2}-5, & x \le 3 \\ \sqrt{x+13}, & x>3. \end{cases}

\text{R H L}=\lim\limits_{x \rightarrow 3^{3}}f(x)=\lim\limits_{h \rightarrow 0}\sqrt{3+h+13}=\sqrt{16}=4

\text{L H L}=\lim\limits_{x \rightarrow 3}f(x)=\lim\limits_{h \rightarrow 0}(3-h)^{2}-5=9-5=4

Since, LHL=RHL, limit exists and is equal to 4.

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