Advertisements
Advertisements
Question
\[\lim_{x \to 1} \frac{\sqrt{5x - 4} - \sqrt{x}}{x^3 - 1}\]
Advertisements
Solution
\[\lim_{x \to 1} \left[ \frac{\sqrt{5x - 4} - \sqrt{x}}{x^3 - 1} \right]\] It is of the form \[\frac{0}{0}\]
Rationalising the numerator:
\[\lim_{x \to 1} \left[ \frac{\left( \sqrt{5x - 4} - \sqrt{x} \right) \left( \sqrt{5x - 4} + \sqrt{x} \right)}{\left( x^3 - 1 \right) \left( \sqrt{5x - 4} + \sqrt{x} \right)} \right]\]
= \[\lim_{x \to 1} \left[ \frac{5x - 4 - x}{\left( x - 1 \right)\left( x^2 + x + 1 \right)\left( \sqrt{5x - 4} + \sqrt{x} \right)} \right]\]
= \[\lim_{x \to 1} \left[ \frac{4\left( x - 1 \right)}{\left( x - 1 \right)\left( x^2 + x + 1 \right)\left( \sqrt{5x - 4} + \sqrt{x} \right)} \right]\]
= \[\frac{4}{3\left( 1 + 1 \right)}\]
= \[\frac{2}{3}\]
APPEARS IN
RELATED QUESTIONS
Find `lim_(x -> 1)` f(x), where `f(x) = {(x^2 -1, x <= 1), (-x^2 -1, x > 1):}`
If f(x) = `{(|x| + 1,x < 0), (0, x = 0),(|x| -1, x > 0):}`
For what value (s) of a does `lim_(x -> a)` f(x) exists?
\[\lim_{x \to 0} \frac{\sqrt{a^2 + x^2} - a}{x^2}\]
\[\lim_{x \to 3} \frac{x - 3}{\sqrt{x - 2} - \sqrt{4 - x}}\]
\[\lim_{x \to 0} \frac{x}{\sqrt{1 + x} - \sqrt{1 - x}}\]
\[\lim_{x \to 1} \frac{x - 1}{\sqrt{x^2 + 3 - 2}}\]
\[\lim_{x \to 3} \frac{\sqrt{x + 3} - \sqrt{6}}{x^2 - 9}\]
\[\lim_{x \to 1} \frac{\sqrt{5x - 4} - \sqrt{x}}{x^2 - 1}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{x}\]
\[\lim_{x \to 2} \frac{\sqrt{x^2 + 1} - \sqrt{5}}{x - 2}\]
\[\lim_{x \to 2} \frac{x - 2}{\sqrt{x} - \sqrt{2}}\]
\[\lim_{x \to 5} \frac{x - 5}{\sqrt{6x - 5} - \sqrt{4x + 5}}\]
\[\lim_{x \to 0} \frac{\sqrt{2 - x} - \sqrt{2 + x}}{x}\]
\[\lim_{x \to 1} \frac{\left( 2x - 3 \right) \left( \sqrt{x} - 1 \right)}{3 x^2 + 3x - 6}\]
\[\lim_{x \to 0} \frac{5^x - 1}{\sqrt{4 + x} - 2}\]
\[\lim_{x \to 0} \frac{a^x + a^{- x} - 2}{x^2}\]
\[\lim_{x \to 0} \frac{a^{mx} - b^{nx}}{x}\]
\[\lim_{x \to 2} \frac{x - 2}{\log_a \left( x - 1 \right)}\]
\[\lim_{x \to 0} \frac{a^x + b^ x - c^x - d^x}{x}\]
\[\lim_{x \to 0} \frac{e^x - 1 + \sin x}{x}\]
\[\lim_{x \to 0} \frac{\sin 2x}{e^x - 1}\]
\[\lim_{x \to 0} \frac{e^{2x} - e^x}{\sin 2x}\]
\[\lim_{x \to 0} \frac{\log \left( 2 + x \right) + \log 0 . 5}{x}\]
`\lim_{x \to \pi/2} \frac{e^\cos x - 1}{\cos x}`
\[\lim_{x \to 0} \frac{e^{3 + x} - \sin x - e^3}{x}\]
\[\lim_{x \to 0} \frac{e^x - x - 1}{2}\]
`\lim_{x \to 0} \frac{e^\tan x - 1}{\tan x}`
`\lim_{x \to 0} \frac{e^\tan x - 1}{x}`
`\lim_{x \to 0} \frac{e^x - e^\sin x}{x - \sin x}`
\[\lim_{x \to 0} \frac{a^x - a^{- x}}{x}\]
\[\lim_{x \to 0} \left\{ \frac{e^x + e^{- x} - 2}{x^2} \right\}^{1/ x^2}\]
Write the value of \[\lim_{x \to \pi/2} \frac{2x - \pi}{\cos x} .\]
Write the value of \[\lim_{n \to \infty} \frac{1 + 2 + 3 + . . . + n}{n^2} .\]
