Advertisements
Advertisements
प्रश्न
Evaluate: `lim_(x -> 1) ((1 + x)^6 - 1)/((1 + x)^2 - 1)`
उत्तर
Given that `lim_(x -> 0) ((1 + x)^6 - 1)/((1 + x)^2 - 1)`
Dividing the numerator and denominator by x, we get
= `lim_(x -> 0) (((1 + x)^6 - 1)/x)/(((1 + x)^2 - 1)/x)`
Putting 1 + x = y
⇒ x = y – 1
= `lim_((y - 1 -> 0),(because y -> 1)) ((y^6 - (1)^6)/(y - 1))/((y^2 - (1)^2)/(y - 1))`
= `(lim_(y -> 1) (y^6 - (1)^6)/(y - 1))/(lim_(y -> 1) (y^2 - (1)^2)/(y - 1))` .....`[lim_(x -> a) (f(x))/(g(x)) = (lim_(x -> a) f(x))/(lim_(x -> a) g(x))]`
= `(6 * (1)^(6 - 1))/(2 * (1)^(2 - 1))`
= `6/2`
= 3 ......`["Using" lim_(x -> a) (x^n - a^n)/(x - a) = n * a^(n - 1)]`
APPEARS IN
संबंधित प्रश्न
\[\lim_{x \to 2} \left( 3 - x \right)\]
\[\lim_{x \to 4} \frac{x^2 - 7x + 12}{x^2 - 3x - 4}\]
\[\lim_{x \to 2} \left( \frac{1}{x - 2} - \frac{2}{x^2 - 2x} \right)\]
\[\lim_{x \to 2} \left( \frac{1}{x - 2} - \frac{4}{x^3 - 2 x^2} \right)\]
\[\lim_{x \to a} \frac{\left( x + 2 \right)^{5/2} - \left( a + 2 \right)^{5/2}}{x - a}\]
\[\lim_{x \to 1} \frac{x^{15} - 1}{x^{10} - 1}\]
\[\lim_{x \to \infty} \frac{3 x^{- 1} + 4 x^{- 2}}{5 x^{- 1} + 6 x^{- 2}}\]
\[\lim_{x \to 0} \frac{7x \cos x - 3 \sin x}{4x + \tan x}\]
\[\lim_{x \to 0} \frac{\cos 3x - \cos 7x}{x^2}\]
\[\lim_{x \to 0} \frac{x \cos x + 2 \sin x}{x^2 + \tan x}\]
\[\lim_{x \to 0} \frac{\sin 5x - \sin 3x}{\sin x}\]
\[\lim_{x \to 0} \frac{\sin 2x \left( \cos 3x - \cos x \right)}{x^3}\]
\[\lim_{x \to 0} \frac{\sin \left( 3 + x \right) - \sin \left( 3 - x \right)}{x}\]
\[\lim_\theta \to 0 \frac{1 - \cos 4\theta}{1 - \cos 6\theta}\]
\[\lim_\theta \to 0 \frac{\sin 4\theta}{\tan 3\theta}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{\cos x} - \sqrt{\sin x}}{x - \frac{\pi}{4}}\]
\[\lim_{x \to a} \frac{\cos x - \cos a}{\sqrt{x} - \sqrt{a}}\]
\[\lim_{x \to \pi} \frac{\sqrt{5 + \cos x} - 2}{\left( \pi - x \right)^2}\]
\[\lim_{x \to 1} \frac{1 - x^2}{\sin \pi x}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{{cosec}^2 x - 2}{\cot x - 1}\]
\[\lim_{x \to \pi} \frac{\sqrt{2 + \cos x} - 1}{\left( \pi - x \right)^2}\]
\[\lim_{x \to 0} \frac{\log \left( a + x \right) - \log a}{x}\]
\[\lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n\]
Write the value of \[\lim_{x \to \infty} \frac{\sin x}{x} .\]
Write the value of \[\lim_{x \to 0^+} \left[ x \right] .\]
\[\lim_{n \to \infty} \left\{ \frac{1}{1 - n^2} + \frac{2}{1 - n^2} + . . . + \frac{n}{1 - n^2} \right\}\]
\[\lim_{x \to \infty} \frac{\sin x}{x}\] equals
The value of \[\lim_{x \to 0} \frac{\sqrt{a^2 - ax + x^2} - \sqrt{a^2 + ax + x^2}}{\sqrt{a + x} - \sqrt{a - x}}\]
Evaluate the following limits: `lim_(x -> 5)[(x^3 - 125)/(x^2 - 25)]`
