Advertisements
Advertisements
Question
If α is a repeated root of ax2 + bx + c = 0, then \[\lim_{x \to \alpha} \frac{\tan \left( a x^2 + bx + c \right)}{\left( x - \alpha \right)^2}\]
Options
a
b
c
0
Advertisements
Solution
\[\lim_{x \to \alpha} \left[ \frac{\tan\left( a x^2 + bx + c \right)}{\left( x - \alpha \right)^2} \right]\]
\[ = \lim_{x \to \alpha} \left[ \frac{\tan\left\{ a\left( x - \alpha \right)\left( x - \alpha \right) \right\}}{a \left( x - \alpha \right)^2} \right] \times a\]
\[ = \lim_{x \to \alpha} \left[ \frac{\tan\left\{ a \left( x - \alpha \right)^2 \right\}}{a \left( x - \alpha \right)^2} \right] \times a\]
\[ = 1 \times a \left[ \lim_\theta \to 0 \left( \frac{\tan\theta}{\theta} \right) = 1 \right]\]
\[ = a\]
APPEARS IN
RELATED QUESTIONS
\[\lim_{x \to 0} \frac{x^{2/3} - 9}{x - 27}\]
\[\lim_{x \to - 1}{\left( 4 x^2 + 2 \right)}\]
\[\lim_{x \to \sqrt{3}} \frac{x^4 - 9}{x^2 + 4\sqrt{3}x - 15}\]
\[\lim_{x \to 3} \left( x^2 - 9 \right) \left[ \frac{1}{x + 3} + \frac{1}{x - 3} \right]\]
\[\lim_{x \to 3} \frac{x^2 - x - 6}{x^3 - 3 x^2 + x - 3}\]
\[\lim_{x \to 1} \left\{ \frac{x - 2}{x^2 - x} - \frac{1}{x^3 - 3 x^2 + 2x} \right\}\]
Evaluate the following limit:
\[\lim_{x \to 1} \frac{x^7 - 2 x^5 + 1}{x^3 - 3 x^2 + 2}\]
\[\lim_{x \to 4} \frac{x^3 - 64}{x^2 - 16}\]
If \[\lim_{x \to 3} \frac{x^n - 3^n}{x - 3} = 108,\] find the value of n.
\[\lim_{x \to \infty} \frac{5 x^3 - 6}{\sqrt{9 + 4 x^6}}\]
\[\lim_{n \to \infty} \left[ \frac{1 + 2 + 3 . . . . . . n - 1}{n^2} \right]\]
\[\lim_{x \to 0} \frac{\tan 8x}{\sin 2x}\]
\[\lim_{x \to 0} \frac{x \cos x + 2 \sin x}{x^2 + \tan x}\]
\[\lim_{x \to 0} \frac{\sin 5x - \sin 3x}{\sin x}\]
\[\lim_{x \to 0} \frac{\sec 5x - \sec 3x}{\sec 3x - \sec x}\]
\[\lim_{x \to 0} \frac{x^2 - \tan 2x}{\tan x}\]
\[\lim_{x \to 0} \frac{5x + 4 \sin 3x}{4 \sin 2x + 7x}\]
\[\lim_{x \to 0} \frac{\tan 2x - \sin 2x}{x^3}\]
Evaluate the following limit:
\[\lim_{x \to 0} \frac{\sin\left( \alpha + \beta \right)x + \sin\left( \alpha - \beta \right)x + \sin2\alpha x}{\cos^2 \beta x - \cos^2 \alpha x}\]
\[\lim_{x \to \frac{\pi}{2}} \frac{\sin 2x}{\cos x}\]
\[\lim_{x \to \frac{\pi}{2}} \frac{\cot x}{\frac{\pi}{2} - x}\]
\[\lim_{x \to a} \frac{\cos x - \cos a}{\sqrt{x} - \sqrt{a}}\]
\[\lim_{x \to 1} \frac{1 + \cos \pi x}{\left( 1 - x \right)^2}\]
\[\lim_{x \to 1} \frac{1 - x^2}{\sin \pi x}\]
\[\lim_{x \to 2} \frac{x^2 - x - 2}{x^2 - 2x + \sin \left( x - 2 \right)}\]
\[\lim_{x \to \pi} \frac{1 + \cos x}{\tan^2 x}\]
Write the value of \[\lim_{x \to 1^-} x - \left[ x \right] .\]
Write the value of \[\lim_{x \to 0^-} \left[ x \right] .\]
Write the value of \[\lim_{x \to 0^-} \frac{\sin \left[ x \right]}{\left[ x \right]} .\]
Write the value of \[\lim_{x \to 0} \frac{\sin x^\circ}{x} .\]
\[\lim_{x \to \infty} \frac{\sin x}{x}\] equals
\[\lim_{x \to 0} \frac{\sin x^0}{x}\]
The value of \[\lim_{n \to \infty} \frac{\left( n + 2 \right)! + \left( n + 1 \right)!}{\left( n + 2 \right)! - \left( n + 1 \right)!}\] is
Evaluate the following limits: `lim_(x -> 2)[(x^(-3) - 2^(-3))/(x - 2)]`
Evaluate the following limit:
`lim_(x -> 7)[((root(3)(x) - root(3)(7))(root(3)(x) + root(3)(7)))/(x - 7)]`
Evaluate the following limits: if `lim_(x -> 5)[(x^"k" - 5^"k")/(x - 5)]` = 500, find all possible values of k.
Evaluate the following limit:
`lim_(x->5)[(x^3-125)/(x^5-3125)]`
Evaluate the following limit:
`lim_(x->7)[((root(3)(x)-root(3)(7))(root(3)(x)+root(3)(7)))/(x-7)]`
Evaluate the following limit.
`lim_(x->5)[(x^3 -125)/(x^5 - 3125)]`
