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Introduction to limits

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Question

If $L = lim_{x \to \infty} \{x - x^{2} \log_{e} (1 + \frac{1}{x})\},$ then the value of 40L is ______ .

Moderate

Solution

Let x = 1/y, Then
$\begin{array}{l} lim_{x \to \infty} (x - x^{2} \log_{e} (1 + \frac{1}{x})) = lim_{y \to 0} (\frac{1}{y} - \frac{\log_{e} (1 + y)}{y^{2}}) \\ = lim_{y \to 0} (\frac{y - \log_{e} (1 + y)}{y^{2}}) \\ = lim_{y \to 0} (\frac{y - (y - \frac{y^{2}}{2})}{y^{2}}) = 1 / 2 \end{array}$

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Similar Questions

Match the following lists:

Column -I	Column -II
A. If $L = lim_{x \to - 1} \frac{\sqrt[3]{(7 - x)} - 2}{(x + 1)}$ , then 12L =	P. -2
B. If $L = lim_{x \to π / 4} \frac{\tan^{3} x - \tan x}{\cos (x + \frac{π}{4})}$ then -L/4=	Q. 2
C. If $L = lim_{x \to 1} \frac{(2 x - 3) (\sqrt{x} - 1)}{2 x^{2} + x - 3}$ , then 20L=	R. 1
D. If $L = lim_{x \to \infty} \frac{\log x^{n} - [x]}{[x]}$ , where $n \in N$ ([x] denotes greatest integer less than or equal to x),then -2L=	S. -1

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