Lim x -1-44 x-1-2 -x-1

Lim_x →1/44x 1/2√(x) 1 It is of the form 0/0 lim_x →1/4 [ (2√(x))^2 1^2/2√(x) 1 ] =lim_x →1/4 [ (2√(x) 1)(2√(x)+1)/(2√(x) 1) ] =lim_x →1/4 (2√(x)+1) =2√(1/4)+1 ...

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Byju's Answer

Standard XII

Mathematics

Expansions to Remove Indeterminate Form

lim x →1/44 x...

Question

${lim}_{x \to \frac{1}{4}} \frac{4 x - 1}{2 \sqrt{x} - 1}$

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Solution

${lim}_{x \to \frac{1}{4}} \frac{4 x - 1}{2 \sqrt{x} - 1}$

It is of the form $\frac{0}{0}$

${lim}_{x \to \frac{1}{4}} [\frac{(2 \sqrt{x})^{2} - 1^{2}}{2 \sqrt{x} - 1}]$

$= {lim}_{x \to \frac{1}{4}} [\frac{(2 \sqrt{x} - 1) (2 \sqrt{x} + 1)}{(2 \sqrt{x} - 1)}]$

$= {lim}_{x \to \frac{1}{4}} (2 \sqrt{x} + 1)$

$= 2 \sqrt{\frac{1}{4}} + 1$

$= 2 (\frac{1}{2}) + 1$

=1+1=2

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Similar questions

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lim x \to 0 \frac{[x]}{x}

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\lim_{x \to 1} \frac{\sqrt{x^{2} - 1} + \sqrt{x - 1}}{\sqrt{x^{2} - 1}}, x > 1

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limx→144x−12√x−1

limx→144x−12√x−1 It is of the form 00 limx→14[(2√x)2−122√x−1] =limx→14[(2√x−1)(2√x+1)(2√x−1)] =limx→14(2√x+1) =2√14+1 =2(12)+1 =1+1=2

${lim}_{x \to \frac{1}{4}} \frac{4 x - 1}{2 \sqrt{x} - 1}$