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Continuity and Differentiability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability1 Mark
Question
Examine the function for continuity.
$f(x)=\frac{x^{2}-25}{x+5}, x \neq-5$
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Answer

The given function is $f(x)=\frac{x^{2}-25}{x+5}, x \neq 5$ 
For any real number $k \neq 5$, we get,
$\lim _{x \rightarrow \mathbf{k}} f(x)=\lim _{x \rightarrow k} \frac{x^{2}-25}{x+5}=\lim _{x \rightarrow k} \frac{(x-5)(x+5)}{x+5}=\lim _{x \rightarrow k}(x-5)=(k-5)$ 
Also, $f(k) = \mathop {\lim }\limits_{x \to k} \frac{{(k - 5)(k + 5)}}{{k + 5}} = \mathop {\lim }\limits_{x \to k} (k - 5) = (k - 5)({\text{ As }}k \ne 5)$ 
Thus, $\lim _{{x} \rightarrow {k}} {f}({x})={f}({k})$ 
Therefore, f is continuous at every point in the domain of f and thus, it is continuous function.

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