lim _{x rightarrow 1} left( lim _{y rightarro | vedclass

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(C) We need to evaluate the limit: $\lim _{x \rightarrow 1} \left( \lim _{y \rightarrow \infty} y \left( e^{x/y} - 1 \right) \right)$.First,consider t

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(C) We need to evaluate the limit: $\lim _{x \rightarrow 1} \left( \lim _{y \rightarrow \infty} y \left( e^{x/y} - 1 \right) \right)$.First,consider the inner limit: $L_{inner} = \lim _{y \rightarrow \infty} y \left( e^{x/y} - 1 \right)$.Using the Taylor series expansion for $e^u = 1 + u + \frac{u^2}{2!} + \dots$,where $u = x/y$:$L_{inner} = \lim _{y \rightarrow \infty} y \left( (1 + \frac{x}{y} + \frac{x^2}{2y^2} + \dots) - 1 \right)$$L_{inner} = \lim _{y \rightarrow \infty} y \left( \frac{x}{y} + \frac{x^2}{2y^2} + \dots \right)$$L_{inner} = \lim _{y \rightarrow \infty} \left( x + \frac{x^2}{2y} + \dots \right) = x$.Now,evaluate the outer limit: $\lim _{x \rightarrow 1} (x) = 1$.

$\lim _{x \rightarrow 1} \left( \lim _{y \rightarrow \infty} y \left( (e^x)^{1/y} - 1 \right) \right) = $

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