If $f(x) = \left\{ \begin{array}{r} \frac{\sin{5x}}{x^{2} + 2x},\ \ \ \ x
eq 0 \ k + \frac{1}{2},\ \ \ x = 0 \end{array} ight.\$ is continuous at $x = 0$, then the value of $k$ is

Question

If $f(x) = \left\{ \begin{array}{r} \frac{\sin{5x}}{x^{2} + 2x},\ \ \ \ x 
eq 0 \ k + \frac{1}{2},\ \ \ x = 0 \end{array} ight.\$ is continuous at $x = 0$, then the value of $k$ is

Dr. Abubacker · Accepted Answer

(c)
LHL$= \lim_{h \rightarrow 0}\ f(0 - h) = \lim_{h \rightarrow 0}\frac{\sin{5(0 - h)}}{{(0 - h)}^{2} + 2(0 - h)}$
$= - \lim_{h \rightarrow 0}\frac{\frac{\sin{5h}}{5h}}{\frac{1}{5}(h - 2)} = \frac{5}{2}$
Since, it is continuous at $x = 0$, therefore LHL$= f(0)$
$\Rightarrow \ \frac{5}{2} = k + \frac{1}{2}\ \ \ \  \Rightarrow \ \ \ \ k = 2$

Q. If $f(x) = \left\{ \begin{array}{r} \frac{\sin{5x}}{x^{2} + 2x},\ \ \ \ x \neq 0 \\ k + \frac{1}{2},\ \ \ x = 0 \end{array} \right.\$ is continuous at $x = 0$, then the value of $k$ is

Explanation

Reviewed by Dr. Abubacker

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