Let k be a non-zero real number. If f(x) =(ex−1)2sin(xk)log(1+x4), x≠0 8 , x=0Is a continuous function at x=0 then the value of k is

∴ limx→0f(x)=limx→0(ex−1)2sin(xk)cos(1+74) f0=limx→0 x2ex−1x24ksin⁡xkxk⋅log⁡1+x4x4 ⇒4k=8⇒k=2

Let k be a non-zero real number. If f(x) =(ex−1)2sin(xk)log(1+x4), x≠0 8 , x=0Is a continuous function at x=0 then the value of k is

∴ limx→0f(x)=limx→0(ex−1)2sin(xk)cos(1+74) f0=limx→0 x2ex−1x24ksin⁡xkxk⋅log⁡1+x4x4 ⇒4k=8⇒k=2

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Let k be a non-zero real number. If $f (x)$ $= \{\begin{cases} \frac{{(e^{x} - 1)}^{2}}{\sin (\frac{x}{k}) \log (1 + \frac{x}{4})}, & x \neq 0 \\ 8, & x = 0 \end{cases}$
Is a continuous function at $x = 0$ then the value of k is

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$∴ \lim_{x \to 0} f (x) = \lim_{x \to 0} \frac{{(e^{x} - 1)}^{2}}{\sin (\frac{x}{k}) \cos (1 + \frac{7}{4})}$

$f (0) \underset{x \to 0}{= lim} \frac{x^{2} {(\frac{e^{x} - 1}{x})}^{2} 4 k}{\frac{\sin \frac{x}{k}}{\frac{x}{k}} \cdot \frac{\log (1 + \frac{x}{4})}{\frac{x}{4}}}$

$\Rightarrow 4 k = 8 \Rightarrow k = 2$

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Let k be a non-zero real number. If f(x) =(ex−1)2sin(xk)log(1+x4), x≠0 8 , x=0Is a continuous function at x=0 then the value of k is

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Let k be a non-zero real number. If $f (x)$ $= \{\begin{cases} \frac{{(e^{x} - 1)}^{2}}{\sin (\frac{x}{k}) \log (1 + \frac{x}{4})}, & x \neq 0 \\ 8, & x = 0 \end{cases}$
Is a continuous function at $x = 0$ then the value of k is