Advertisements
Advertisements
प्रश्न
\[\lim_{x \to \sqrt{10}} \frac{\sqrt{7 + 2x} - \left( \sqrt{5} + \sqrt{2} \right)}{x^2 - 10}\]
उत्तर
\[\lim_{x \to \sqrt{10}} \left[ \frac{\sqrt{7 + 2x} - \left( \sqrt{5} + \sqrt{2} \right)}{x^2 - \left( \sqrt{10} \right)^2} \right]\]
= \[\lim_{x \to \sqrt{10}} \left[ \frac{\sqrt{7 + 2x} - \sqrt{\left( \sqrt{5} + \sqrt{2} \right)^2}}{\left( x - \sqrt{10} \right)\left( x + \sqrt{10} \right)} \right]\]
= \[\lim_{x \to \sqrt{10}} \left[ \frac{\sqrt{7 + 2x} - \sqrt{5 + 2 + 2\sqrt{5}\sqrt{2}}}{\left( x - \sqrt{10} \right)\left( x + \sqrt{10} \right)} \right]\]
= \[\lim_{x \to \sqrt{10}} \left[ \frac{\sqrt{7 + 2x} - \sqrt{7 + 2\sqrt{10}}}{\left( x - \sqrt{10} \right)\left( x + \sqrt{10} \right)} \right]\]
Rationalising the numerator:
\[\lim_{x \to \sqrt{10}} \left[ \frac{\left( \sqrt{7 + 2x} - \sqrt{7 + 2\sqrt{10}} \right) \left( \sqrt{7 + 2x} + \sqrt{7 + 2\sqrt{10}} \right)}{\left( x - \sqrt{10} \right)\left( x + \sqrt{10} \right) \left( \sqrt{7 + 2x} + \sqrt{7 + 2\sqrt{10}} \right)} \right]\]
= \[\lim_{x \to \sqrt{10}} \left[ \frac{\left( 7 + 2x \right) - \left( 7 + 2\sqrt{10} \right)}{\left( x - \sqrt{10} \right)\left( x + \sqrt{10} \right) \left( \sqrt{7 + 2x} + \sqrt{7 + 2\sqrt{10}} \right)} \right]\]
= \[\lim_{x \to \sqrt{10}} \left[ \frac{2\left( x - \sqrt{10} \right)}{\left( x - \sqrt{10} \right)\left( x + \sqrt{10} \right) \left( \sqrt{7 + 2x} + \sqrt{7 + 2\sqrt{10}} \right)} \right]\]
= \[\frac{2}{\left( \sqrt{10} + \sqrt{10} \right) \left( 2\sqrt{7 + 2\sqrt{10}} \right)}\]
\[= \frac{2}{\left( 2\sqrt{10} \right) \times 2\sqrt{7 + 2\sqrt{10}}}\]
\[ = \frac{1}{2\sqrt{10}\sqrt{\left( \sqrt{5} + \sqrt{2} \right)^2}}\]
= \[\frac{1}{2\sqrt{10}\left( \sqrt{5} + \sqrt{2} \right)} \times \frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} - \sqrt{2}}\]
\[= \frac{1}{2\sqrt{10}}\left[ \frac{\sqrt{5} - \sqrt{2}}{\left( \sqrt{5} \right)^2 - \left( \sqrt{2} \right)^2} \right]\]
= \[\frac{\sqrt{5} - \sqrt{2}}{6\sqrt{10}}\]
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):}`
\[\lim_{x \to 0} \frac{\sqrt{1 + x} - \sqrt{1 - x}}{2x}\]
\[\lim_{x \to 3} \frac{x - 3}{\sqrt{x - 2} - \sqrt{4 - x}}\]
\[\lim_{x \to 1} \frac{\sqrt{5x - 4} - \sqrt{x}}{x - 1}\]
\[\lim_{x \to 1} \frac{x - 1}{\sqrt{x^2 + 3 - 2}}\]
\[\lim_{x \to 1} \frac{\sqrt{5x - 4} - \sqrt{x}}{x^2 - 1}\]
\[\lim_{x \to 2} \frac{\sqrt{x^2 + 1} - \sqrt{5}}{x - 2}\]
\[\lim_{x \to 2} \frac{x - 2}{\sqrt{x} - \sqrt{2}}\]
\[\lim_{x \to 0} \frac{\sqrt{a + x} - \sqrt{a}}{x\sqrt{a^2 + ax}}\]
\[\lim_{x \to 5} \frac{x - 5}{\sqrt{6x - 5} - \sqrt{4x + 5}}\]
\[\lim_{x \to 1} \frac{\sqrt{5x - 4} - \sqrt{x}}{x^3 - 1}\]
\[\lim_{x \to 0} \frac{\sqrt{2 - x} - \sqrt{2 + x}}{x}\]
\[\lim_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h}, x \neq 0\]
\[\lim_{x \to \sqrt{6}} \frac{\sqrt{5 + 2x} - \left( \sqrt{3} + \sqrt{2} \right)}{x^2 - 6}\]
\[\lim_{x \to 0} \frac{5^x - 1}{\sqrt{4 + x} - 2}\]
\[\lim_{x \to 0} \frac{8^x - 4^x - 2^x + 1}{x^2}\]
\[\lim_{x \to 0} \frac{a^x + b^x + c^x - 3}{x}\]
\[\lim_{x \to 0} \frac{5^x + 3^x + 2^x - 3}{x}\]
\[\lim_{x \to \infty} \left( a^{1/x} - 1 \right)x\]
\[\lim_{x \to 0} \frac{a^{mx} - b^{nx}}{\sin kx}\]
\[\lim_{x \to a} \frac{\log x - \log a}{x - a}\]
\[\lim_{x \to 0} \frac{\log \left( 2 + x \right) + \log 0 . 5}{x}\]
\[\lim_{x \to 0} \frac{\log \left| 1 + x^3 \right|}{\sin^3 x}\]
\[\lim_{x \to 0} \frac{e^x - 1}{\sqrt{1 - \cos x}}\]
\[\lim_{x \to 0} \frac{e^{3 + x} - \sin x - e^3}{x}\]
\[\lim_{x \to 0} \frac{e^{3x} - e^{2x}}{x}\]
`\lim_{x \to 0} \frac{e^\tan x - 1}{x}`
\[\lim_{x \to \infty} \left\{ \frac{x^2 + 2x + 3}{2 x^2 + x + 5} \right\}^\frac{3x - 2}{3x + 2}\]
\[\lim_{x \to 0} \left\{ \frac{e^x + e^{- x} - 2}{x^2} \right\}^{1/ x^2}\]
\[\lim_{x \to a} \left\{ \frac{\sin x}{\sin a} \right\}^\frac{1}{x - a}\]
Write the value of \[\lim_{x \to - \infty} \left( 3x + \sqrt{9 x^2 - x} \right) .\]
Evaluate: `lim_(x -> 2) (x^2 - 4)/(sqrt(3x - 2) - sqrt(x + 2))`
