Advertisements
Advertisements
Question
\[\lim_{x \to \sqrt{10}} \frac{\sqrt{7 + 2x} - \left( \sqrt{5} + \sqrt{2} \right)}{x^2 - 10}\]
Advertisements
Solution
\[\lim_{x \to \sqrt{10}} \left[ \frac{\sqrt{7 + 2x} - \left( \sqrt{5} + \sqrt{2} \right)}{x^2 - \left( \sqrt{10} \right)^2} \right]\]
= \[\lim_{x \to \sqrt{10}} \left[ \frac{\sqrt{7 + 2x} - \sqrt{\left( \sqrt{5} + \sqrt{2} \right)^2}}{\left( x - \sqrt{10} \right)\left( x + \sqrt{10} \right)} \right]\]
= \[\lim_{x \to \sqrt{10}} \left[ \frac{\sqrt{7 + 2x} - \sqrt{5 + 2 + 2\sqrt{5}\sqrt{2}}}{\left( x - \sqrt{10} \right)\left( x + \sqrt{10} \right)} \right]\]
= \[\lim_{x \to \sqrt{10}} \left[ \frac{\sqrt{7 + 2x} - \sqrt{7 + 2\sqrt{10}}}{\left( x - \sqrt{10} \right)\left( x + \sqrt{10} \right)} \right]\]
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
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)`.
\[\lim_{x \to 2} \frac{\sqrt{3 - x} - 1}{2 - x}\]
\[\lim_{x \to 1} \frac{x - 1}{\sqrt{x^2 + 3 - 2}}\]
\[\lim_{x \to 3} \frac{\sqrt{x + 3} - \sqrt{6}}{x^2 - 9}\]
\[\lim_{x \to 1} \frac{\sqrt{5x - 4} - \sqrt{x}}{x^2 - 1}\]
\[\lim_{x \to 5} \frac{x - 5}{\sqrt{6x - 5} - \sqrt{4x + 5}}\]
\[\lim_{x \to 2} \frac{\sqrt{1 + 4x} - \sqrt{5 + 2x}}{x - 2}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + 3x} - \sqrt{1 - 3x}}{x}\]
\[\lim_{x \to 0} \frac{\sqrt{2 - x} - \sqrt{2 + x}}{x}\]
\[\lim_{x \to \sqrt{6}} \frac{\sqrt{5 + 2x} - \left( \sqrt{3} + \sqrt{2} \right)}{x^2 - 6}\]
\[\lim_{x \to \sqrt{2}} \frac{\sqrt{3 + 2x} - \left( \sqrt{2} + 1 \right)}{x^2 - 2}\]
\[\lim_{x \to 0} \frac{\log \left( 1 + x \right)}{3^x - 1}\]
\[\lim_{x \to 0} \frac{a^x + a^{- x} - 2}{x^2}\]
\[\lim_{x \to 0} \frac{a^x + b^x - 2}{x}\]
\[\lim_{x \to 2} \frac{x - 2}{\log_a \left( x - 1 \right)}\]
\[\lim_{x \to 0} \frac{\sin 2x}{e^x - 1}\]
\[\lim_{x \to 0} \frac{x\left( 2^x - 1 \right)}{1 - \cos x}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{\log \left( 1 + x \right)}\]
\[\lim_{x \to 0} \frac{\log \left| 1 + x^3 \right|}{\sin^3 x}\]
`\lim_{x \to \pi/2} \frac{a^\cot x - a^\cos x}{\cot x - \cos x}`
\[\lim_{x \to 0} \frac{e^x - 1}{\sqrt{1 - \cos x}}\]
`\lim_{x \to \pi/2} \frac{e^\cos x - 1}{\cos x}`
\[\lim_{x \to 0} \frac{e^x - x - 1}{2}\]
`\lim_{x \to 0} \frac{e^\tan x - 1}{\tan x}`
`\lim_{x \to 0} \frac{e^\tan x - 1}{x}`
`\lim_{x \to 0} \frac{e^x - e^\sin x}{x - \sin x}`
\[\lim_{x \to 0} \frac{3^{2 + x} - 9}{x}\]
\[\lim_{x \to \infty} \left\{ \frac{x^2 + 2x + 3}{2 x^2 + x + 5} \right\}^\frac{3x - 2}{3x + 2}\]
\[\lim_{x \to 1} \left\{ \frac{x^3 + 2 x^2 + x + 1}{x^2 + 2x + 3} \right\}^\frac{1 - \cos \left( x - 1 \right)}{\left( x - 1 \right)^2}\]
\[\lim_{x \to 0} \left\{ \frac{e^x + e^{- x} - 2}{x^2} \right\}^{1/ x^2}\]
Write the value of \[\lim_{n \to \infty} \frac{n! + \left( n + 1 \right)!}{\left( n + 1 \right)! + \left( n + 2 \right)!} .\]
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 ______.
