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
Find `lim_(x -> 5) f(x)`, where f(x) = |x| - 5
\[\lim_{x \to - 5} \frac{2 x^2 + 9x - 5}{x + 5}\]
\[\lim_{x \to 3} \frac{x^4 - 81}{x^2 - 9}\]
\[\lim_{x \to 4} \frac{x^2 - 16}{\sqrt{x} - 2}\]
\[\lim_{x \to 1} \left( \frac{1}{x - 1} - \frac{2}{x^2 - 1} \right)\]
\[\lim_{x \to a} \frac{\left( x + 2 \right)^{5/2} - \left( a + 2 \right)^{5/2}}{x - a}\]
\[\lim_{x \to 0} \frac{\left( 1 + x \right)^6 - 1}{\left( 1 + x \right)^2 - 1}\]
If \[\lim_{x \to 3} \frac{x^n - 3^n}{x - 3} = 108,\] find the value of n.
If \[\lim_{x \to a} \frac{x^9 - a^9}{x - a} = \lim_{x \to 5} \left( 4 + x \right),\] find all possible values of a.
\[\lim_{n \to \infty} \left[ \frac{1^3 + 2^3 + . . . n^3}{\left( n - 1 \right)^4} \right]\]
\[\lim_{x \to - \infty} \left( \sqrt{4 x^2 - 7x} + 2x \right)\]
\[\lim_{x \to 0} \frac{\tan mx}{\tan nx}\]
\[\lim_{x \to 0} \frac{\sin \left( 2 + x \right) - \sin \left( 2 - x \right)}{x}\]
\[\lim_{x \to 0} \frac{1 - \cos 2x}{\cos 2x - \cos 8x}\]
\[\lim_{x \to 0} \frac{\sqrt{2} - \sqrt{1 + \cos x}}{x^2}\]
\[\lim_{x \to 0} \frac{\cos 2x - 1}{\cos x - 1}\]
Evaluate the following limits:
\[\lim_{x \to 0} \frac{2\sin x - \sin2x}{x^3}\]
\[\lim_{x \to 0} \left( cosec x - \cot x \right)\]
If \[\lim_{x \to 0} kx cosec x = \lim_{x \to 0} x cosec kx,\]
\[\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2} - \cos x - \sin x}{\left( \frac{\pi}{4} - x \right)^2}\]
\[\lim_{x \to a} \frac{\cos x - \cos a}{\sqrt{x} - \sqrt{a}}\]
\[\lim_{x \to \pi} \frac{\sqrt{5 + \cos x} - 2}{\left( \pi - x \right)^2}\]
\[\lim_{x \to a} \frac{\sin \sqrt{x} - \sin \sqrt{a}}{x - a}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{f\left( x \right) - f\left( \frac{\pi}{4} \right)}{x - \frac{\pi}{4}},\]
\[\lim_{x \to \frac{\pi}{6}} \frac{\cot^2 x - 3}{cosec x - 2}\]
\[\lim_{x \to \pi} \frac{1 + \cos x}{\tan^2 x}\]
\[\lim_{x \to 0} \left( \cos x \right)^{1/\sin x}\]
Write the value of \[\lim_{x \to \infty} \frac{\sin x}{x} .\]
Write the value of \[\lim_{x \to 1^-} x - \left[ x \right] .\]
\[\lim_{x \to \infty} \frac{\sqrt{x^2 - 1}}{2x + 1}\]
\[\lim_{h \to 0} \left\{ \frac{1}{h\sqrt[3]{8 + h}} - \frac{1}{2h} \right\} =\]
\[\lim_{x \to 0} \frac{8}{x^8}\left\{ 1 - \cos \frac{x^2}{2} - \cos \frac{x^2}{4} + \cos \frac{x^2}{2} \cos \frac{x^2}{4} \right\}\] is equal to
The value of \[\lim_{x \to \infty} \frac{n!}{\left( n + 1 \right)! - n!}\]
The value of \[\lim_{x \to \infty} \frac{\left( x + 1 \right)^{10} + \left( x + 2 \right)^{10} + . . . + \left( x + 100 \right)^{10}}{x^{10} + {10}^{10}}\] is
The value of \[\lim_{n \to \infty} \left\{ \frac{1 + 2 + 3 + . . . + n}{n + 2} - \frac{n}{2} \right\}\]
\[\lim_{x \to 0} \frac{\left| \sin x \right|}{x}\]
If \[f\left( x \right) = \begin{cases}\frac{\sin\left[ x \right]}{\left[ x \right]}, & \left[ x \right] \neq 0 \\ 0, & \left[ x \right] = 0\end{cases}\] where denotes the greatest integer function, then \[\lim_{x \to 0} f\left( x \right)\]
If `f(x) = {{:(x + 2",", x ≤ - 1),(cx^2",", x > -1):}`, find 'c' if `lim_(x -> -1) f(x)` exists
Evaluate the following limits: `lim_(x -> 3) [sqrt(x + 6)/x]`
