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

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Property 4

lim x → 15 x-...

Question

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

Open in App

Solution

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

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

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

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

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

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

= $\frac{4}{2}$

= 2

Suggest Corrections

Similar questions

\lim_{x \to 1} \frac{\sqrt{5 x - 4} - \sqrt{x}}{x^{3} - 1}

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

Q. Arrange the following limits in the ascending order :
(1)

{lim}_{x \to \infty} {(\frac{1 + x}{2 + x})}^{x + 2}

(2)

{lim}_{x \to 0} (1 + 2 x)^{3 / x}

(3)

{lim}_{θ \to 0} \frac{sin θ}{2 θ}

(4)

{lim}_{x \to 0} \frac{{log}_{e} (1 + x)}{x}

$lim x \to \infty (1 + \frac{1}{x})^{x} =$

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

Join BYJU'S Learning Program

limx→1 5x-4-xx-1

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

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