Advertisements
Advertisements
प्रश्न
Evaluate the following limits: `lim_(x -> 0)[((1 - x)^8 - 1)/((1 - x)^2 - 1)]`
Advertisements
उत्तर १
`lim_(x -> 0)((1 - x)^8 - 1)/((1 - x)^2 - 1)`
Put 1 – x = y
As x → 0, y → 1
∴ `lim_(x -> 0)((1 - x)^8 - 1)/((1 - x)^2 - 1) = lim_(y -> 1)(y^8 - 1^8)/(y^2 - 1^2)`
= `lim_(y -> 1)((y^8 - 1^8)/(y - 1))/((y^2 - 1^2)/(y - 1)) ...[(because y -> 1"," therefore y ≠ 1","),(therefore y - 1 ≠ 0)]`
= `(lim_(y -> 1)(y^8 - 1^8)/(y - 1))/(lim_(y -> 1)(y^2 - 1^2)/(y - 1))`
= `(8(1)^7)/(2(1)^1) ...[because lim_(x -> "a")(x^"n" - "a"^"n")/(x - "a") = "na"^("n" - 1)]`
= 4
उत्तर २
`lim_(x -> 0) ((1 - x)^8 - 1)/((1 - x)^2 - 1)`
Put 1 – x = y
As x → 0, y → 1
∴ `lim_(x -> 0)((1 - x)^8 - 1)/((1 - x)^2 - 1) = lim_(y -> 1)(y^8 - 1)/(y^2 - 1)`
= `lim_(y -> 1) ((y^4 - 1)(y^4 + 1))/(y^2 - 1)`
= `lim_(y -> 1) ((y^2 - 1)(y^2 + 1)(y^4 + 1))/(y^2 - 1)`
= `lim_(y -> 1)(y^2 + 1) (y^4 + 1) ...[(because y -> 1 therefore y ≠ 1),(therefore y^2 ≠ 1),(therefore y^2 - 1 ≠ 0)]`
= (2) (2) = 4
APPEARS IN
संबंधित प्रश्न
Suppose f(x) = `{(a+bx, x < 1),(4, x = 1),(b-ax, x > 1):}` and if `lim_(x -> 1) f(x) = f(1)` what are possible values of a and b?
\[\lim_{x \to 1} \frac{\sqrt{x + 8}}{\sqrt{x}}\]
\[\lim_{x \to 1/4} \frac{4x - 1}{2\sqrt{x} - 1}\]
\[\lim_{x \to 1} \frac{\sqrt{x^2 - 1} + \sqrt{x - 1}}{\sqrt{x^2 - 1}}, x > 1\]
\[\lim_{x \to \infty} \sqrt{x^2 + cx - x}\]
\[\lim_{x \to 0} \frac{\sin 3x}{5x}\]
\[\lim_{x \to \frac{\pi}{3}} \frac{\sqrt{3} - \tan x}{\pi - 3x}\]
\[\lim_{x \to a} \frac{\cos \sqrt{x} - \cos \sqrt{a}}{x - a}\]
\[\lim_{x \to \infty} \left\{ \frac{3 x^2 + 1}{4 x^2 - 1} \right\}^\frac{x^3}{1 + x}\]
\[\lim_{n \to \infty} \frac{n!}{\left( n + 1 \right)! + n!}\] is equal to
\[\lim_{x \to 2} \frac{\sqrt{1 + \sqrt{2 + x} - \sqrt{3}}}{x - 2}\] is equal to
If α is a repeated root of ax2 + bx + c = 0, then \[\lim_{x \to \alpha} \frac{\tan \left( a x^2 + bx + c \right)}{\left( x - \alpha \right)^2}\]
If the value of `lim_(x -> 1) (1 - (1 - x))^"m"/x` is 99, then n = ______.
Evaluate the following limit :
`lim_(x->3)[sqrt(x+6)/x]`
Evaluate the following limit :
`lim_(x->7)[[(root3(x)- root3(7))(root3(x) + root3(7)))/(x-7)]`
Evaluate the following limits: `lim_(x -> 3) [sqrt(x + 6)/x]`
Evaluate the following limit.
`lim_(x->3)[sqrt(x + 6)/x]`
