Advertisements
Advertisements
प्रश्न
\[\lim_{x \to 1} \frac{\sqrt{3 + x} - \sqrt{5 - x}}{x^2 - 1}\]
Advertisements
उत्तर
\[\lim_{x \to 1} \left[ \frac{\sqrt{3 + x} - \sqrt{5 - x}}{x^2 - 1} \right]\] It is of the form \[\frac{0}{0}\]
Rationalising the numerator:
\[\lim_{x \to 1} \left[ \frac{\left( \sqrt{3 + x} - \sqrt{5 - x} \right)\left( \sqrt{3 + x} + \sqrt{5 - x} \right)}{\left( x - 1 \right)\left( x + 1 \right)\left( \sqrt{3 + x} + \sqrt{5 - x} \right)} \right]\]
= \[\lim_{x \to 1} \left[ \frac{\left( 3 + x \right) - \left( 5 - x \right)}{\left( x - 1 \right)\left( x + 1 \right)\left\{ \sqrt{3 + x} + \sqrt{5 - x} \right\}} \right]\]
= \[\lim_{x \to 1} \left[ \frac{2x - 2}{\left( x - 1 \right)\left( x + 1 \right) \left\{ \sqrt{3 + x} + \sqrt{5 - x} \right\}} \right]\]
= \[\lim_{x \to 1} \left[ \frac{2\left( x - 1 \right)}{\left( x - 1 \right)\left( x + 1 \right)\left\{ \sqrt{3 + x} + \sqrt{5 - x} \right\}} \right]\]
= \[\frac{2}{2 \left( \sqrt{4} + \sqrt{4} \right)}\]
= \[\frac{1}{4}\]
APPEARS IN
संबंधित प्रश्न
Find `lim_(x -> 0)` f(x) and `lim_(x -> 1)` f(x) where f(x) = `{(2x + 3, x <= 0),(3(x+1), x > 0):}`
Evaluate `lim_(x -> 0) f(x)` where `f(x) = { (|x|/x, x != 0),(0, x = 0):}`
if `f(x) = { (mx^2 + n, x < 0),(nx + m, 0<= x <= 1),(nx^3 + m, x > 1):}`
For what integers m and n does `lim_(x-> 0) f(x)` and `lim_(x -> 1) f(x)` exist?
\[\lim_{x \to 0} \frac{\sqrt{a^2 + x^2} - a}{x^2}\]
\[\lim_{x \to 3} \frac{x - 3}{\sqrt{x - 2} - \sqrt{4 - x}}\]
\[\lim_{x \to 0} \frac{x}{\sqrt{1 + x} - \sqrt{1 - x}}\]
\[\lim_{x \to 1} \frac{\sqrt{5x - 4} - \sqrt{x}}{x^2 - 1}\]
\[\lim_{x \to 2} \frac{x - 2}{\sqrt{x} - \sqrt{2}}\]
\[\lim_{x \to 0} \frac{\sqrt{a + x} - \sqrt{a}}{x\sqrt{a^2 + ax}}\]
\[\lim_{x \to 5} \frac{x - 5}{\sqrt{6x - 5} - \sqrt{4x + 5}}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + x^2} - \sqrt{1 - x^2}}{x}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + x^2} - \sqrt{1 + x}}{\sqrt{1 + x^3} - \sqrt{1 + x}}\]
\[\lim_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h}, x \neq 0\]
\[\lim_{x \to \sqrt{10}} \frac{\sqrt{7 + 2x} - \left( \sqrt{5} + \sqrt{2} \right)}{x^2 - 10}\]
\[\lim_{x \to \sqrt{6}} \frac{\sqrt{5 + 2x} - \left( \sqrt{3} + \sqrt{2} \right)}{x^2 - 6}\]
\[\lim_{x \to 0} \frac{a^{mx} - 1}{b^{nx} - 1}, n \neq 0\]
\[\lim_{x \to 0} \frac{a^x + b^x - 2}{x}\]
\[\lim_{x \to \infty} \left( a^{1/x} - 1 \right)x\]
\[\lim_{x \to 0} \frac{a^{mx} - b^{nx}}{\sin kx}\]
\[\lim_{x \to 5} \frac{e^x - e^5}{x - 5}\]
\[\lim_{x \to 0} \frac{e^{x + 2} - e^2}{x}\]
\[\lim_{x \to 0} \frac{e^{3 + x} - \sin x - e^3}{x}\]
\[\lim_{x \to 0} \frac{e^x - x - 1}{2}\]
`\lim_{x \to 0} \frac{e^\tan x - 1}{x}`
\[\lim_{x \to 0} \frac{3^{2 + x} - 9}{x}\]
\[\lim_{x \to 0} \frac{x\left( e^x - 1 \right)}{1 - \cos x}\]
\[\lim_{x \to \pi/2} \frac{2^{- \cos x} - 1}{x\left( x - \frac{\pi}{2} \right)}\]
\[\lim_{x \to a} \left\{ \frac{\sin x}{\sin a} \right\}^\frac{1}{x - a}\]
\[\lim_{x \to \infty} \left\{ \frac{3 x^2 + 1}{4 x^2 - 1} \right\}^\frac{x^3}{1 + x}\]
Write the value of \[\lim_{x \to \pi/2} \frac{2x - \pi}{\cos x} .\]
