Dear Student, Please find below the solution to the asked query: We haveL=limx→∞ x ∑i=1n e1x-1As sumation variable is i, hence we can take e1x-1 out of sumation=limx→∞xe1x-1 ∑i=1n 1 =limx→∞xe1x-1 1+1+1+1 n times=limx→∞xe1x-1 n=n limx→∞xe1x-1=n limx→∞ e1x-11xAs limx→∞1x=0, hence, limx→∞ e1x-11x=1⇒n limx→∞ e1x-11x=n×1=nlimx→∞ x ∑i=1n e1x-1=n Hope this information will clear your doubts about this topic If you have any doubts just ask here on the ask and answer forum and our experts will try to help you out as soon as possible Regards

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Please find below the solution to the asked query:

We have L = \lim_{x \to \infty} \{x \sum_{i = 1}^{n} (e^{\frac{1}{x}} - 1)\} As sumation variable is i, hence we can take (e^{\frac{1}{x}} - 1) out of sumation = \lim_{x \to \infty} \{x (e^{\frac{1}{x}} - 1) \sum_{i = 1}^{n} 1\} = \lim_{x \to \infty} \{x (e^{\frac{1}{x}} - 1) (1 + 1 + 1 + 1 . . . . n times)\} = \lim_{x \to \infty} \{x (e^{\frac{1}{x}} - 1) n\} = n \lim_{x \to \infty} x (e^{\frac{1}{x}} - 1) = n \lim_{x \to \infty} \frac{e^{\frac{1}{x}} - 1}{\frac{1}{x}} As \lim_{x \to \infty} \frac{1}{x} = 0, hence, \lim_{x \to \infty} \frac{e^{\frac{1}{x}} - 1}{\frac{1}{x}} = 1 \Rightarrow n \lim_{x \to \infty} \frac{e^{\frac{1}{x}} - 1}{\frac{1}{x}} = n \times 1 = n \lim_{x \to \infty} \{x \sum_{i = 1}^{n} (e^{\frac{1}{x}} - 1)\} = n

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