Advertisements
Advertisements
Question
\[\lim_{x \to 2} \frac{x - 2}{\sqrt{x} - \sqrt{2}}\]
Solution
\[\lim_{x \to 2} \left[ \frac{x - 2}{\sqrt{x} - \sqrt{2}} \right]\] It is of the form \[\frac{0}{0}\]
⇒ \[\lim_{x \to 2} \left[ \frac{\left( \sqrt{x} \right)^2 - \left( \sqrt{2} \right)^2}{x - \sqrt{2}} \right]\]
= \[\lim_{x \to 2} \left[ \frac{\left( \sqrt{x} - \sqrt{2} \right)\left( \sqrt{x} + \sqrt{2} \right)}{\left( x - \sqrt{2} \right)} \right]\]
= \[\sqrt{2} + \sqrt{2}\]
= \[2\sqrt{2}\]
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?
If the function f(x) satisfies `lim_(x -> 1) (f(x) - 2)/(x^2 - 1) = pi`, evaluate `lim_(x -> 1) f(x)`.
\[\lim_{x \to 0} \frac{2x}{\sqrt{a + x} - \sqrt{a - x}}\]
\[\lim_{x \to 3} \frac{\sqrt{x + 3} - \sqrt{6}}{x^2 - 9}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{x}\]
\[\lim_{x \to 0} \frac{\sqrt{a + x} - \sqrt{a}}{x\sqrt{a^2 + ax}}\]
\[\lim_{x \to 5} \frac{x - 5}{\sqrt{6x - 5} - \sqrt{4x + 5}}\]
\[\lim_{x \to 1} \frac{\sqrt{5x - 4} - \sqrt{x}}{x^3 - 1}\]
\[\lim_{x \to 2} \frac{\sqrt{1 + 4x} - \sqrt{5 + 2x}}{x - 2}\]
\[\lim_{x \to 4} \frac{2 - \sqrt{x}}{4 - x}\]
\[\lim_{x \to 1} \frac{\sqrt{3 + x} - \sqrt{5 - x}}{x^2 - 1}\]
\[\lim_{x \to 1} \frac{\left( 2x - 3 \right) \left( \sqrt{x} - 1 \right)}{3 x^2 + 3x - 6}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + x^2} - \sqrt{1 + x}}{\sqrt{1 + x^3} - \sqrt{1 + x}}\]
\[\lim_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h}, x \neq 0\]
\[\lim_{x \to \sqrt{6}} \frac{\sqrt{5 + 2x} - \left( \sqrt{3} + \sqrt{2} \right)}{x^2 - 6}\]
\[\lim_{x \to 0} \frac{5^x - 1}{\sqrt{4 + x} - 2}\]
\[\lim_{x \to 0} \frac{a^x + a^{- x} - 2}{x^2}\]
\[\lim_{x \to 0} \frac{a^{mx} - b^{nx}}{x}\]
\[\lim_{x \to a} \frac{\log x - \log a}{x - a}\]
\[\lim_{x \to 0} \frac{x\left( 2^x - 1 \right)}{1 - \cos x}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{\log \left( 1 + x \right)}\]
\[\lim_{x \to 0} \frac{\log \left| 1 + x^3 \right|}{\sin^3 x}\]
\[\lim_{x \to 0} \frac{e^x - 1}{\sqrt{1 - \cos x}}\]
\[\lim_{x \to 0} \frac{e^{3x} - e^{2x}}{x}\]
`\lim_{x \to 0} \frac{e^x - e^\sin x}{x - \sin x}`
\[\lim_{x \to 0} \frac{3^{2 + x} - 9}{x}\]
\[\lim_{x \to 0} \frac{x\left( e^x - 1 \right)}{1 - \cos x}\]
\[\lim_{x \to \pi/2} \frac{2^{- \cos x} - 1}{x\left( x - \frac{\pi}{2} \right)}\]
\[\lim_{x \to 0} \left\{ \frac{e^x + e^{- x} - 2}{x^2} \right\}^{1/ x^2}\]
\[\lim_{x \to 0} \left\{ \frac{e^x + e^{- x} - 2}{x^2} \right\}^{1/ x^2}\]
\[\lim_{x \to a} \left\{ \frac{\sin x}{\sin a} \right\}^\frac{1}{x - a}\]
Write the value of \[\lim_{x \to \pi/2} \frac{2x - \pi}{\cos x} .\]
Write the value of \[\lim_{n \to \infty} \frac{1 + 2 + 3 + . . . + n}{n^2} .\]
Evaluate: `lim_(h -> 0) (sqrt(x + h) - sqrt(x))/h`
Let f(x) be a polynomial of degree 4 having extreme values at x = 1 and x = 2. If `lim_(x rightarrow 0) ((f(x))/x^2 + 1)` = 3 then f(–1) is equal to ______.