Let k in mathbb{R}. If lim _{x rightarrow 0^{ | vedclass

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(B) The given limit is of the form $1^\infty$. We use the formula $\lim_{x 	o a} f(x)^{g(x)} = e^{\lim_{x 	o a} g(x)(f(x)-1)}$.Given $\lim _{x igh

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$2$

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(B) The given limit is of the form $1^\infty$. We use the formula $\lim_{x 	o a} f(x)^{g(x)} = e^{\lim_{x 	o a} g(x)(f(x)-1)}$.Given $\lim _{x ightarrow 0^{+}}(\sin (\sin k x)+\cos x+x)^{\frac{2}{x}}= e ^6$,we have:$e^{\lim _{x ightarrow 0^{+}} \frac{2}{x}(\sin (\sin k x)+\cos x+x-1)} = e^6$.Equating the exponents:$\lim _{x ightarrow 0^{+}} \frac{2}{x}(\sin (\sin k x) + (\cos x - 1) + x) = 6$.$\lim _{x}$ ${ightarrow 0^{+}} \left[ 2 \cdot \frac{\sin(\sin kx)}{\sin kx} \cdot \frac{\sin kx}{kx} \cdot k + 2 \cdot \frac{\cos x - 1}{x^2} \cdot x + 2 \cdot \frac{x}{x} ight] = 6$.Using standard limits $\lim_{u 	o 0} \frac{\sin u}{u} = 1$ and $\lim_{x 	o 0} \frac{1-\cos x}{x^2} = \frac{1}{2}$:$2(1 \cdot 1 \cdot k) + 2(-\frac{1}{2} \cdot 0) + 2(1) = 6$.$2k + 2 = 6$ $\Rightarrow 2k = 4$ $\Rightarrow k = 2$.

Let $k \in \mathbb{R}$. If $\lim _{x \rightarrow 0^{+}}(\sin (\sin k x)+\cos x+x)^{\frac{2}{x}}= e ^6$,then the value of $k$ is

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