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1 votes
1 votes
Q1> $\lim_{x->0} xlogx$

Q2> $\lim_{x\rightarrow Inf} \frac{sin x}{ x}$

for first question why we can't take:

$\frac{x^{2} log x}{x}$

1 Answer

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2 votes

Q.1. $\lim_{x\rightarrow 0}$ xlogx = $\lim_{x\rightarrow 0}$ $\frac{log x}{\frac{1}{x}}$  $\frac{infinity}{infinity}$ form. Use L'hospital rule

= $\lim_{x\rightarrow 0}$ $\frac{\frac{1}{x}}{\frac{-1}{x^{2}}}$

= $\lim_{x\rightarrow 0}$ -x

= 0

Q.2. $\lim_{x\rightarrow Inf}$ $\frac{sinx}{x}$ = 0

Here, sin x is a bounded function i.e. its range is fixed [-1,1] whatever may be the value of x.

So, $\lim_{x\rightarrow Inf}$ $\frac{sinx}{x}$ will give a finite value divided by an infinite value which is 0.

edited by

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