Advertisements
Advertisements
प्रश्न
\[\lim_{x \to 1} \frac{\left( 2x - 3 \right) \left( \sqrt{x} - 1 \right)}{3 x^2 + 3x - 6}\]
Advertisements
उत्तर
\[\lim_{x \to 1} \left[ \frac{\left( 2x - 3 \right)\left( \sqrt{x} - 1 \right)}{3 x^2 + 3x - 6} \right]\] It is of the form \[\frac{0}{0}\]
⇒ \[\lim_{x \to 1} \left[ \frac{\left( 2x - 3 \right)\left( \sqrt{x} - 1 \right)}{3\left( x^2 + x - 2 \right)} \right]\]
= \[\lim_{x \to 1} \left[ \frac{\left( 2x - 3 \right)\left( \sqrt{x} - 1 \right)}{3\left( x^2 + 2x - x - 2 \right)} \right]\]
= \[\lim_{x \to 1} \left[ \frac{\left( 2x - 3 \right)\left( \sqrt{x} - 1 \right)}{3\left( x\left( x + 2 \right) - 1\left( x + 2 \right) \right)} \right]\]
= \[\lim_{x \to 1} \left[ \frac{\left( 2x - 3 \right)\left( \sqrt{x} - 1 \right)}{3\left( x - 1 \right)\left( x + 2 \right)} \right]\]
= \[\lim_{x \to 1} \left[ \frac{\left( 2x - 3 \right)\left( \sqrt{x} - 1 \right)}{3\left( \left( \sqrt{x} \right)^2 - 1^2 \right)\left( x + 2 \right)} \right]\]
= \[\lim_{x \to 1} \left[ \frac{\left( 2x - 3 \right)\left( \sqrt{x} - 1 \right)}{3\left( \sqrt{x} + 1 \right)\left( \sqrt{x} - 1 \right)\left( x + 2 \right)} \right]\]
= \[\frac{- 1}{3\left( 2 \right) \times 3}\]
= \[\frac{- 1}{18}\]
APPEARS IN
संबंधित प्रश्न
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)`.
\[\lim_{x \to 0} \frac{2x}{\sqrt{a + x} - \sqrt{a - x}}\]
\[\lim_{x \to 0} \frac{\sqrt{a^2 + x^2} - a}{x^2}\]
\[\lim_{x \to 0} \frac{x}{\sqrt{1 + x} - \sqrt{1 - x}}\]
\[\lim_{x \to 2} \frac{\sqrt{x^2 + 1} - \sqrt{5}}{x - 2}\]
\[\lim_{x \to 2} \frac{x - 2}{\sqrt{x} - \sqrt{2}}\]
\[\lim_{x \to 0} \frac{\sqrt{a + x} - \sqrt{a}}{x\sqrt{a^2 + ax}}\]
\[\lim_{x \to 5} \frac{x - 5}{\sqrt{6x - 5} - \sqrt{4x + 5}}\]
\[\lim_{x \to 1} \frac{\sqrt{5x - 4} - \sqrt{x}}{x^3 - 1}\]
\[\lim_{x \to 2} \frac{\sqrt{1 + 4x} - \sqrt{5 + 2x}}{x - 2}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + 3x} - \sqrt{1 - 3x}}{x}\]
\[\lim_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h}, x \neq 0\]
\[\lim_{x \to 0} \frac{a^x + b^x - 2}{x}\]
\[\lim_{x \to 0} \frac{9^x - 2 . 6^x + 4^x}{x^2}\]
\[\lim_{x \to 0} \frac{8^x - 4^x - 2^x + 1}{x^2}\]
\[\lim_{x \to 0} \frac{a^x + b^x + c^x - 3}{x}\]
\[\lim_{x \to 0} \frac{5^x + 3^x + 2^x - 3}{x}\]
\[\lim_{x \to 0} \frac{e^x - 1 + \sin x}{x}\]
\[\lim_{x \to 0} \frac{e\sin x - 1}{x}\]
\[\lim_{x \to a} \frac{\log x - \log a}{x - a}\]
\[\lim_{x \to 0} \frac{\log \left( a + x \right) - \log \left( a - x \right)}{x}\]
\[\lim_{x \to 0} \frac{\log \left( 2 + x \right) + \log 0 . 5}{x}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{\log \left( 1 + x \right)}\]
\[\lim_{x \to 0} \frac{\log \left| 1 + x^3 \right|}{\sin^3 x}\]
`\lim_{x \to \pi/2} \frac{e^\cos x - 1}{\cos x}`
`\lim_{x \to 0} \frac{e^\tan x - 1}{\tan x}`
\[\lim_{x \to 0} \frac{e^{bx} - e^{ax}}{x} \text{ where } 0 < a < b\]
\[\lim_{x \to 0} \frac{3^{2 + x} - 9}{x}\]
\[\lim_{x \to a} \left\{ \frac{\sin x}{\sin a} \right\}^\frac{1}{x - a}\]
\[\lim_{x \to \infty} \left\{ \frac{3 x^2 + 1}{4 x^2 - 1} \right\}^\frac{x^3}{1 + x}\]
\[\lim_{x \to 0} \left\{ \frac{e^x + e^{- x} - 2}{x^2} \right\}^{1/ x^2}\]
Evaluate: `lim_(h -> 0) (sqrt(x + h) - sqrt(x))/h`
