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# Find the value of $k$ so that the value of function$f$ is continuous at the indicated points.$f(x) = \left\{ {\begin{array}{*{20}{c}} {\dfrac{{k\cos x}}{{\pi - 2x}},x \ne \dfrac{\pi }{2}} \\ {3,x = \dfrac{\pi }{2}} \end{array}} \right.{\text{ at }}x = \dfrac{\pi }{2}$ . Verified
A function ${\text{f(x)}}$ is continuous at ${\text{x = c}}$ , if
$\mathop {\lim }\limits_{x \to 0} f(x) = f({\text{c}})$
Here, $f\left( {\dfrac{\pi }{2}} \right) = 3$
$\therefore {\text{ }}\mathop {\lim }\limits_{x \to \dfrac{\pi }{2}} \dfrac{{{\text{k}}\cos x}}{{\pi - 2x}} = 3$
Since, left side of the equation is of $\dfrac{0}{0}$ form so,
$\Rightarrow \mathop {\lim }\limits_{x \to \dfrac{\pi }{2}} \dfrac{{ - {\text{k}}\sin x}}{{ - 2}} = 3 \\ \Rightarrow \dfrac{{{\text{k}}\sin \dfrac{\pi }{2}}}{2} = 3 \\ \Rightarrow {\text{k = 6}} \\$
Hence, for ${\text{k = 6}}$, ${\text{f(x)}}$ will be continuous at ${\text{x = }}\dfrac{\pi }{2}$
Note – As we know L-Hospital’s rule is applicable in limits if and only if the function under limit is of $\dfrac{0}{0}$ form or of $\dfrac{\infty }{\infty }$ form. In such cases all we need to do is differentiate the numerator and denominator separately and further continue with the limit. In the above question, the same was the case. As the function was present in $\dfrac{0}{0}$ form, so we have used L-Hospital’s rule.