Advertisements
Advertisements
Question
\[\lim_{x \to 1} \frac{ x^2 - \sqrt{x}}{\sqrt{x} - 1}\]
Advertisements
Solution
\[\lim_{x \to 1} \left[ \frac{x^2 - \sqrt{x}}{\sqrt{x} - 1} \right]\]It is of the form \[\frac{0}{0}\]
\[\lim_{x \to 1} \left[ \frac{\sqrt{x}\left( x\sqrt{x} - 1 \right)}{\sqrt{x} - 1} \right]\]
= \[\lim_{x \to 1} \left[ \frac{\sqrt{x}\left( x^{3/2} - 1 \right)}{x^\frac{1}{2} - 1} \right]\]
= \[\lim_{x \to 1} \left[ \frac{\sqrt{x}\left( \left( \sqrt{x} \right)^3 - 1^3 \right)}{\left( \sqrt{x} - 1 \right)} \right] \left[ A^3 - B^3 = \left( A - B \right)\left( A^2 + AB + B^2 \right) \right]\]
=\[\lim_{x \to 1} \left[ \frac{\left( \sqrt{x} \right)\left( \sqrt{x} - 1 \right)\left( x + \sqrt{x} + 1 \right)}{\left( \sqrt{x} - 1 \right)} \right]\]
= 1 (1 + 1 + 1)
= 3
APPEARS IN
RELATED QUESTIONS
\[\lim_{x \to 0} \frac{\sqrt{1 + x + x^2} - 1}{x}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + x} - \sqrt{1 - x}}{2x}\]
\[\lim_{x \to 1} \frac{\sqrt{5x - 4} - \sqrt{x}}{x - 1}\]
\[\lim_{x \to 1} \frac{x - 1}{\sqrt{x^2 + 3 - 2}}\]
\[\lim_{x \to 3} \frac{\sqrt{x + 3} - \sqrt{6}}{x^2 - 9}\]
\[\lim_{x \to 2} \frac{\sqrt{x^2 + 1} - \sqrt{5}}{x - 2}\]
\[\lim_{x \to 7} \frac{4 - \sqrt{9 + x}}{1 - \sqrt{8 - x}}\]
\[\lim_{x \to 0} \frac{\sqrt{a + x} - \sqrt{a}}{x\sqrt{a^2 + ax}}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + x^2} - \sqrt{1 - x^2}}{x}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + x + x^2} - \sqrt{x + 1}}{2 x^2}\]
\[\lim_{x \to 4} \frac{2 - \sqrt{x}}{4 - x}\]
\[\lim_{x \to 1} \frac{\sqrt{3 + x} - \sqrt{5 - x}}{x^2 - 1}\]
\[\lim_{x \to \sqrt{6}} \frac{\sqrt{5 + 2x} - \left( \sqrt{3} + \sqrt{2} \right)}{x^2 - 6}\]
\[\lim_{x \to 0} \frac{\log \left( 1 + x \right)}{3^x - 1}\]
\[\lim_{x \to 0} \frac{a^x + a^{- x} - 2}{x^2}\]
\[\lim_{x \to 0} \frac{8^x - 4^x - 2^x + 1}{x^2}\]
\[\lim_{x \to 0} \frac{a^x + b^x + c^x - 3}{x}\]
\[\lim_{x \to 0} \frac{5^x + 3^x + 2^x - 3}{x}\]
\[\lim_{x \to 0} \frac{a^{mx} - b^{nx}}{\sin kx}\]
\[\lim_{x \to 0} \frac{\sin 2x}{e^x - 1}\]
\[\lim_{x \to 0} \frac{e\sin x - 1}{x}\]
\[\lim_{x \to a} \frac{\log x - \log a}{x - a}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{\log \left( 1 + x \right)}\]
\[\lim_{x \to 0} \frac{\log \left| 1 + x^3 \right|}{\sin^3 x}\]
\[\lim_{x \to 0} \frac{e^x - 1}{\sqrt{1 - \cos x}}\]
\[\lim_{x \to 0} \frac{e^{x + 2} - e^2}{x}\]
\[\lim_{x \to 0} \frac{e^{bx} - e^{ax}}{x} \text{ where } 0 < a < b\]
`\lim_{x \to 0} \frac{e^\tan x - 1}{x}`
\[\lim_{x \to 1} \left\{ \frac{x^3 + 2 x^2 + x + 1}{x^2 + 2x + 3} \right\}^\frac{1 - \cos \left( x - 1 \right)}{\left( x - 1 \right)^2}\]
\[\lim_{x \to 0} \left\{ \frac{e^x + e^{- x} - 2}{x^2} \right\}^{1/ x^2}\]
\[\lim_{x \to \infty} \left\{ \frac{3 x^2 + 1}{4 x^2 - 1} \right\}^\frac{x^3}{1 + x}\]
Write the value of \[\lim_{x \to - \infty} \left( 3x + \sqrt{9 x^2 - x} \right) .\]
Evaluate: `lim_(h -> 0) (sqrt(x + h) - sqrt(x))/h`
Evaluate: `lim_(x -> 2) (x^2 - 4)/(sqrt(3x - 2) - sqrt(x + 2))`
