Advertisements
Advertisements
Question
\[\lim_{x \to 5} \frac{x - 5}{\sqrt{6x - 5} - \sqrt{4x + 5}}\]
Advertisements
Solution
\[\lim_{x \to 5} \left[ \frac{x - 5}{\sqrt{6x - 5} - \sqrt{4x + 5}} \right]\] It is of the form \[\frac{0}{0}\]
Rationalising the denominator:
\[\lim_{x \to 5} \left[ \frac{\left( x - 5 \right) \left( \sqrt{6x - 5} + \sqrt{4x + 5} \right)}{\left( \sqrt{6x - 5} - \sqrt{4x + 5} \right) \left( \sqrt{6x - 5} + \sqrt{4x + 5} \right)} \right]\]
= \[\lim_{x \to 5} \left[ \frac{\left( x - 5 \right) \left( \sqrt{6x - 5} + \sqrt{4x + 5} \right)}{\left( 6x - 5 \right) - \left( 4x + 5 \right)} \right]\]
= \[\lim_{x \to 5} \left[ \frac{\left( x - 5 \right) \left( \sqrt{6x - 5} + \sqrt{4x + 5} \right)}{2x - 10} \right]\]
=\[\lim_{x \to 5} \left[ \frac{\left( x - 5 \right)\left( \sqrt{6x - 5} + \sqrt{4x + 5} \right)}{2\left( x - 5 \right)} \right]\]
= \[\frac{\sqrt{6 \times 5 - 5} + \sqrt{4 \times 5 + 5}}{2}\]
= \[\frac{5 + 5}{2}\]
= 5
APPEARS IN
RELATED QUESTIONS
Evaluate `lim_(x -> 0) f(x)` where `f(x) = { (|x|/x, x != 0),(0, x = 0):}`
Find `lim_(x -> 0)` f(x), where `f(x) = {(x/|x|, x != 0),(0, x = 0):}`
If the function f(x) satisfies `lim_(x -> 1) (f(x) - 2)/(x^2 - 1) = pi`, evaluate `lim_(x -> 1) f(x)`.
\[\lim_{x \to 0} \frac{\sqrt{1 + x + x^2} - 1}{x}\]
\[\lim_{x \to 0} \frac{2x}{\sqrt{a + x} - \sqrt{a - x}}\]
\[\lim_{x \to 2} \frac{\sqrt{3 - x} - 1}{2 - x}\]
\[\lim_{x \to 3} \frac{x - 3}{\sqrt{x - 2} - \sqrt{4 - x}}\]
\[\lim_{x \to 0} \frac{x}{\sqrt{1 + x} - \sqrt{1 - x}}\]
\[\lim_{x \to 1} \frac{\sqrt{5x - 4} - \sqrt{x}}{x - 1}\]
\[\lim_{x \to 1} \frac{x - 1}{\sqrt{x^2 + 3 - 2}}\]
\[\lim_{x \to 3} \frac{\sqrt{x + 3} - \sqrt{6}}{x^2 - 9}\]
\[\lim_{x \to 0} \frac{\sqrt{a + x} - \sqrt{a}}{x\sqrt{a^2 + ax}}\]
\[\lim_{x \to 1} \frac{\sqrt{5x - 4} - \sqrt{x}}{x^3 - 1}\]
\[\lim_{x \to 2} \frac{\sqrt{1 + 4x} - \sqrt{5 + 2x}}{x - 2}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + 3x} - \sqrt{1 - 3x}}{x}\]
\[\lim_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h}, x \neq 0\]
\[\lim_{x \to 0} \frac{5^x - 1}{\sqrt{4 + x} - 2}\]
\[\lim_{x \to 0} \frac{\log \left( 1 + x \right)}{3^x - 1}\]
\[\lim_{x \to 0} \frac{a^x + a^{- x} - 2}{x^2}\]
\[\lim_{x \to 0} \frac{9^x - 2 . 6^x + 4^x}{x^2}\]
\[\lim_{x \to 0} \frac{8^x - 4^x - 2^x + 1}{x^2}\]
\[\lim_{x \to 0} \frac{a^{mx} - b^{nx}}{x}\]
\[\lim_{x \to 0} \frac{e^x - 1 + \sin x}{x}\]
\[\lim_{x \to a} \frac{\log x - \log a}{x - a}\]
\[\lim_{x \to 0} \frac{x\left( 2^x - 1 \right)}{1 - \cos x}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{\log \left( 1 + x \right)}\]
\[\lim_{x \to 0} \frac{e^x - 1}{\sqrt{1 - \cos x}}\]
`\lim_{x \to \pi/2} \frac{e^\cos x - 1}{\cos x}`
\[\lim_{x \to 0} \frac{3^{2 + x} - 9}{x}\]
\[\lim_{x \to \pi/2} \frac{2^{- \cos x} - 1}{x\left( x - \frac{\pi}{2} \right)}\]
\[\lim_{x \to \infty} \left\{ \frac{x^2 + 2x + 3}{2 x^2 + x + 5} \right\}^\frac{3x - 2}{3x + 2}\]
\[\lim_{x \to 0} \left\{ \frac{e^x + e^{- x} - 2}{x^2} \right\}^{1/ x^2}\]
\[\lim_{x \to a} \left\{ \frac{\sin x}{\sin a} \right\}^\frac{1}{x - a}\]
\[\lim_{x \to 0} \frac{\sin x}{\sqrt{1 + x} - 1} .\]
Evaluate: `lim_(x -> 2) (x^2 - 4)/(sqrt(3x - 2) - sqrt(x + 2))`
