Advertisements
Advertisements
प्रश्न
\[\lim_{x \to 4} \frac{x^2 - 16}{\sqrt{x} - 2}\]
Advertisements
उत्तर
\[\lim_{x \to 4} \left[ \frac{x^2 - 16}{\sqrt{x} - 2} \right]\]
\[\text{ It is of the form } \frac{0}{0} . \]
\[ \lim_{x \to 4} \left[ \frac{x^2 - 4^2}{\sqrt{x} - 2} \right]\]
\[ = \lim_{x \to 4} \left[ \frac{\left( x - 4 \right)\left( x + 4 \right)}{\left( \sqrt{x} - 2 \right)} \right]\]
\[ = \lim_{x \to 4} \left[ \frac{\left\{ \left( \sqrt{x} \right)^2 - 2^2 \right\}\left( x + 4 \right)}{\left( \sqrt{x} - 2 \right)} \right]\]
\[ = \lim_{x \to 4} \left[ \frac{\left( \sqrt{x} - 2 \right)\left( \sqrt{x} + 2 \right)\left( x + 4 \right)}{\left( \sqrt{x} - 2 \right)} \right]\]
\[ = \left( 2 + 2 \right)\left( 4 + 4 \right)\]
\[ = 32\]
APPEARS IN
संबंधित प्रश्न
Find `lim_(x -> 5) f(x)`, where f(x) = |x| - 5
\[\lim_{x \to 0} \frac{2 x^2 + 3x + 4}{x^2 + 3x + 2}\]
\[\lim_{x \to 0} \frac{ax + b}{cx + d}, d \neq 0\]
\[\lim_{x \to 2} \left( \frac{x}{x - 2} - \frac{4}{x^2 - 2x} \right)\]
\[\lim_{x \to 1} \left( \frac{1}{x^2 + x - 2} - \frac{x}{x^3 - 1} \right)\]
\[\lim_{x \to \sqrt{3}} \frac{x^4 - 9}{x^2 + 4\sqrt{3}x - 15}\]
\[\lim_{x \to 0} \frac{\left( 1 + x \right)^6 - 1}{\left( 1 + x \right)^2 - 1}\]
\[\lim_{n \to \infty} \frac{n^2}{1 + 2 + 3 + . . . + n}\]
\[\lim_{x \to 0} \frac{\sin 3x}{5x}\]
\[\lim_{x \to 0} \frac{\sin 3x - \sin x}{\sin x}\]
\[\lim_{x \to 0} \frac{\sin \left( 2 + x \right) - \sin \left( 2 - x \right)}{x}\]
\[\lim_{h \to 0} \frac{\left( a + h \right)^2 \sin \left( a + h \right) - a^2 \sin a}{h}\]
\[\lim_{x \to 0} \frac{\sqrt{2} - \sqrt{1 + \cos x}}{x^2}\]
\[\lim_{x \to 0} \frac{3 \sin^2 x - 2 \sin x^2}{3 x^2}\]
\[\lim_{x \to 0} \frac{x \cos x + \sin x}{x^2 + \tan x}\]
\[\lim_{x \to 0} \frac{1 - \cos 2x}{3 \tan^2 x}\]
\[\lim_{x \to \frac{\pi}{8}} \frac{\cot 4x - \cos 4x}{\left( \pi - 8x \right)^3}\]
\[\lim_{x \to \pi} \frac{\sqrt{5 + \cos x} - 2}{\left( \pi - x \right)^2}\]
\[\lim_{x \to a} \frac{\cos \sqrt{x} - \cos \sqrt{a}}{x - a}\]
\[\lim_{x \to a} \frac{\sin \sqrt{x} - \sin \sqrt{a}}{x - a}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{f\left( x \right) - f\left( \frac{\pi}{4} \right)}{x - \frac{\pi}{4}},\]
\[\lim_{x \to \frac{\pi}{4}} \frac{1 - \tan x}{1 - \sqrt{2} \sin x}\]
\[\lim_{x \to \frac{\pi}{2}} \left( \frac{\pi}{2} - x \right) \tan x\]
\[\lim_{x \to \pi} \frac{\sqrt{2 + \cos x} - 1}{\left( \pi - x \right)^2}\]
\[\lim_{x \to 0} \frac{\sin 2x}{e^x - 1}\]
\[\lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n\]
\[\lim_{x \to 0} \left( \cos x + \sin x \right)^{1/x}\]
\[\lim_{x \to \pi} \frac{\sin x}{x - \pi} .\]
Write the value of \[\lim_{x \to 0^+} \left[ x \right] .\]
\[\lim_{x \to 0} \frac{\sin x^0}{x}\]
\[\lim_{x \to a} \frac{x^n - a^n}{x - a}\] 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
\[\lim_{x \to \pi/3} \frac{\sin \left( \frac{\pi}{3} - x \right)}{2 \cos x - 1}\] is equal to
\[\lim_{x \to \pi/4} \frac{4\sqrt{2} - \left( \cos x + \sin x \right)^5}{1 - \sin 2x}\] is equal to
Evaluate the following Limits: `lim_(x -> "a") ((x + 2)^(5/3) - ("a" + 2)^(5/3))/(x - "a")`
Evaluate the following limit :
`lim_(x->7)[[(root3(x)- root3(7))(root3(x) + root3(7)))/(x-7)]`
