Advertisements
Advertisements
Question
\[\lim_{x \to \sqrt{2}} \frac{x^2 - 2}{x^2 + \sqrt{2}x - 4}\]
Advertisements
Solution
\[\lim_{x \to \sqrt{2}} \left[ \frac{x^2 - 2}{x^2 + \sqrt{2}x - 4} \right]\]
\[\text{ It is of the form } \frac{0}{0} . \]
\[ \lim_{x \to \sqrt{2}} \left[ \frac{x^2 - \left( \sqrt{2} \right)^2}{x^2 + 2\sqrt{2}x - \sqrt{2}x - 4} \right]\]
\[ = \lim_{x \to \sqrt{2}} \left[ \frac{\left( x - \sqrt{2} \right)\left( x + \sqrt{2} \right)}{x\left( x + 2\sqrt{2} \right) - \sqrt{2}\left( x + 2\sqrt{2} \right)} \right]\]
\[ = \lim_{x \to \sqrt{2}} \left[ \frac{\left( x - \sqrt{2} \right)\left( x + \sqrt{2} \right)}{\left( x - \sqrt{2} \right)\left( x + 2\sqrt{2} \right)} \right]\]
\[ = \frac{\sqrt{2} + \sqrt{2}}{\sqrt{2} + 2\sqrt{2}}\]
\[ = \frac{2\sqrt{2}}{3\sqrt{2}}\]
\[ = \frac{2}{3}\]
APPEARS IN
RELATED QUESTIONS
\[\lim_{x \to 3} \frac{\sqrt{2x + 3}}{x + 3}\]
\[\lim_{x \to 4} \frac{x^2 - 7x + 12}{x^2 - 3x - 4}\]
\[\lim_{x \to 5} \frac{x^2 - 9x + 20}{x^2 - 6x + 5}\]
\[\lim_{x \to \sqrt{3}} \frac{x^4 - 9}{x^2 + 4\sqrt{3}x - 15}\]
\[\lim_{x \to \sqrt{3}} \frac{x^4 - 9}{x^2 + 4\sqrt{3}x - 15}\]
\[\lim_{x \to 2} \frac{x^3 + 3 x^2 - 9x - 2}{x^3 - x - 6}\]
\[\lim_{x \to 1} \frac{1 - x^{- 1/3}}{1 - x^{- 2/3}}\]
\[\lim_{x \to 1} \frac{x^{15} - 1}{x^{10} - 1}\]
\[\lim_{x \to - 1} \frac{x^3 + 1}{x + 1}\]
\[\lim_{x \to \infty} \left[ \left\{ \sqrt{x + 1} - \sqrt{x} \right\} \sqrt{x + 2} \right]\]
\[\lim_{x \to \infty} \left[ \frac{x^4 + 7 x^3 + 46x + a}{x^4 + 6} \right]\] where a is a non-zero real number.
Show that \[\lim_{x \to \infty} \left( \sqrt{x^2 + x + 1} - x \right) \neq \lim_{x \to \infty} \left( \sqrt{x^2 + 1} - x \right)\]
\[\lim_{x \to 0} \frac{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}{x}\]
\[\lim_{x \to 0} \frac{1 - \cos 4x}{x^2}\]
Evaluate the following limits:
\[\lim_{x \to 0} \frac{2\sin x - \sin2x}{x^3}\]
\[\lim_{x \to 0} \frac{\sin ax + bx}{ax + \sin bx}\]
If \[\lim_{x \to 0} kx cosec x = \lim_{x \to 0} x cosec kx,\]
\[\lim_{x \to \pi} \frac{\sin x}{\pi - x}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{\cos x} - \sqrt{\sin x}}{x - \frac{\pi}{4}}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2} - \cos x - \sin x}{\left( \frac{\pi}{4} - x \right)^2}\]
\[\lim_{x \to \pi} \frac{\sqrt{5 + \cos x} - 2}{\left( \pi - x \right)^2}\]
\[\lim_{x \to 1} \frac{1 - x^2}{\sin \pi x}\]
\[\lim_{x \to 2} \frac{x^2 - x - 2}{x^2 - 2x + \sin \left( x - 2 \right)}\]
\[\lim_{x \to 0} \left( \cos x + a \sin bx \right)^{1/x}\]
\[\lim_{x \to \infty} \frac{\sin x}{x} .\]
Write the value of \[\lim_{x \to 2} \frac{\left| x - 2 \right|}{x - 2} .\]
\[\lim_{x \to 0} \frac{\left( 1 - \cos 2x \right) \sin 5x}{x^2 \sin 3x} =\]
\[\lim_{x \to 0} \frac{x}{\tan x} is\]
\[\lim_{x \to 3} \frac{x - 3}{\left| x - 3 \right|},\] is equal to
\[\lim_{x \to a} \frac{x^n - a^n}{x - a}\] is equal at
\[\lim_{x \to 1} \frac{\sin \pi x}{x - 1}\]
The value of \[\lim_{x \to \infty} \frac{\sqrt{1 + x^4} + \left( 1 + x^2 \right)}{x^2}\] is
\[\lim_{x \to 3} \frac{\sum^n_{r = 1} x^r - \sum^n_{r = 1} 3^r}{x - 3}\]is real to
\[\lim_{x \to \infty} a^x \sin \left( \frac{b}{a^x} \right), a, b > 1\] is equal to
The value of \[\lim_{x \to 0} \frac{1 - \cos x + 2 \sin x - \sin^3 x - x^2 + 3 x^4}{\tan^3 x - 6 \sin^2 x + x - 5 x^3}\] is
if `lim_(x -> 2) (x^"n"- 2^"n")/(x - 2)` = 80 then find the value of n.
Let `f(x) = {{:((k cos x)/(pi - 2x)",", "when" x ≠ pi/2),(3",", x = pi/2 "and if" f(x) = f(pi/2)):}` find the value of k.
