The value of k (k 0),for which the function | vedclass

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(C) For the function $f(x)$ to be continuous at $x = 0$,we must have $\lim_{x 	o 0} f(x) = f(0) = 8$. Evaluating the limit: $\lim_{x 	o 0} \frac{(e^

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$2$

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(C) For the function $f(x)$ to be continuous at $x = 0$,we must have $\lim_{x 	o 0} f(x) = f(0) = 8$. Evaluating the limit: $\lim_{x 	o 0} \frac{(e^x - 1)^4}{\sin(\frac{x^2}{k^2}) \log(1 + \frac{x^2}{2})} = 8$. Dividing the numerator and denominator by $x^4$,we get: $\lim_{x 	o 0} \frac{(\frac{e^x - 1}{x})^4}{\frac{\sin(\frac{x^2}{k^2})}{x^2} \cdot \frac{\log(1 + \frac{x^2}{2})}{x^2}} = 8$. Using standard limits $\lim_{u 	o 0} \frac{e^u - 1}{u} = 1$,$\lim_{u 	o 0} \frac{\sin u}{u} = 1$,and $\lim_{u 	o 0} \frac{\log(1 + u)}{u} = 1$: $\lim_{x 	o 0} \frac{(\frac{e^x - 1}{x})^4}{\frac{\sin(\frac{x^2}{k^2})}{\frac{x^2}{k^2}} \cdot \frac{1}{k^2} \cdot \frac{\log(1 + \frac{x^2}{2})}{\frac{x^2}{2}} \cdot \frac{1}{2}} = 8$. Substituting the limits: $\frac{1^4}{1 \cdot \frac{1}{k^2} \cdot 1 \cdot \frac{1}{2}} = 8$. $\frac{1}{\frac{1}{2k^2}} = 2k^2 = 8$. $k^2 = 4$,which gives $k = 2$ (since $k > 0$).

The value of $k$ $(k > 0)$,for which the function $f(x) = \frac{(e^x - 1)^4}{\sin(\frac{x^2}{k^2}) \log(1 + \frac{x^2}{2})}$,where $x \neq 0$ and $f(0) = 8$,is continuous at $x = 0$,is

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