If a, b, c and k are non-zero real numbers an | vedclass

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(B) Given $\lim _{x \rightarrow \infty} x\left(a^{1 / x}+b^{1 / x}+c^{1 / x}-3 k^{1 / x}\right)=0$. Let $y = \frac{1}{x}$. As $x \rightarrow \infty$,$

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$(abc)^{1/3}$

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(B) Given $\lim _{x \rightarrow \infty} x\left(a^{1 / x}+b^{1 / x}+c^{1 / x}-3 k^{1 / x}\right)=0$. Let $y = \frac{1}{x}$. As $x \rightarrow \infty$,$y \rightarrow 0$. The expression becomes $\lim _{y \rightarrow 0} \frac{a^y+b^y+c^y-3 k^y}{y} = 0$. This can be rewritten as $\lim _{y \rightarrow 0} \left[ \frac{a^y-1}{y} + \frac{b^y-1}{y} + \frac{c^y-1}{y} - 3 \frac{k^y-1}{y} \right] = 0$. Using the standard limit $\lim _{y \rightarrow 0} \frac{a^y-1}{y} = \ln a$,we get: $\ln a + \ln b + \ln c - 3 \ln k = 0$. $\ln (abc) = 3 \ln k$. $\ln (abc) = \ln (k^3)$. Therefore,$k^3 = abc$,which implies $k = (abc)^{1/3}$.

If $a, b, c$ and $k$ are non-zero real numbers and $\lim _{x \rightarrow \infty} x\left(a^{\frac{1}{x}}+b^{\frac{1}{x}}+c^{\frac{1}{x}}-3 k^{\frac{1}{x}}\right)=0$,then $k=$

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