Lim x - 15 x-4-x x 2-1

Limx #8594;15x 4 xx2 1 It is of the form 00. Rationalising the numerator: nbsp;limx #8594;15x 4 x5x 4+x5x 4+xx2 1 = nbsp;limx #8594;15x 4 x5x 4+xx ...

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Byju's Answer

Standard XII

Mathematics

Property 4

lim x → 15 x-...

Question

$\lim_{x \to 1} \frac{\sqrt{5 x - 4} - \sqrt{x}}{x^{2} - 1}$

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Solution

$\lim_{x \to 1} [\frac{\sqrt{5 x - 4} - \sqrt{x}}{x^{2} - 1}]$

It is of the form $\frac{0}{0}$ .
Rationalising the numerator:

$\lim_{x \to 1} [\frac{(\sqrt{5 x - 4} - \sqrt{x}) (\sqrt{5 x - 4} + \sqrt{x})}{(\sqrt{5 x - 4} + \sqrt{x}) (x^{2} - 1)}]$

= $\lim_{x \to 1} [\frac{5 x - 4 - x}{(\sqrt{5 x - 4} + \sqrt{x}) (x - 1) (x + 1)}]$

= $\lim_{x \to 1} [\frac{4 (x - 1)}{(\sqrt{5 x - 4} + \sqrt{x}) (x - 1) (x + 1)}]$

= $\frac{4}{(\sqrt{5 - 4} + \sqrt{1}) (1 + 1)}$

= $\frac{4}{2 \times 2}$

= 1

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limx→1 5x-4-xx2-1

limx→15x-4-xx2-1 It is of the form 00. Rationalising the numerator: limx→15x-4-x5x-4+x5x-4+xx2-1 = limx→15x-4-x5x-4+xx-1x+1 = limx→14x-15x-4+xx-1x+1 = 45-4+11+1 = 42×2 = 1

$\lim_{x \to 1} \frac{\sqrt{5 x - 4} - \sqrt{x}}{x^{2} - 1}$