English

If α is a Repeated Root of Ax2 + Bx + C = 0, Then Lim X → α Tan ( a X 2 + B X + C ) ( X − α ) 2 - Mathematics

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

  •  

  •  

  •  0

MCQ
Advertisements

Solution

\[\lim_{x \to \alpha} \left[ \frac{\tan\left( a x^2 + bx + c \right)}{\left( x - \alpha \right)^2} \right]\]
\[ = \lim_{x \to \alpha} \left[ \frac{\tan\left\{ a\left( x - \alpha \right)\left( x - \alpha \right) \right\}}{a \left( x - \alpha \right)^2} \right] \times a\]
\[ = \lim_{x \to \alpha} \left[ \frac{\tan\left\{ a \left( x - \alpha \right)^2 \right\}}{a \left( x - \alpha \right)^2} \right] \times a\]
\[ = 1 \times a \left[ \lim_\theta \to 0 \left( \frac{\tan\theta}{\theta} \right) = 1 \right]\]
\[ = a\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 29: Limits - Exercise 29.13 [Page 80]

APPEARS IN

RD Sharma Mathematics [English] Class 11
Chapter 29 Limits
Exercise 29.13 | Q 30 | Page 80

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

\[\lim_{x \to 1} \frac{1 + \left( x - 1 \right)^2}{1 + x^2}\]


\[\lim_{x \to 0} \frac{3x + 1}{x + 3}\] 


\[\lim_{x \to \sqrt{3}} \frac{x^4 - 9}{x^2 + 4\sqrt{3}x - 15}\]


\[\lim_{x \to - 2} \frac{x^3 + x^2 + 4x + 12}{x^3 - 3x + 2}\]


\[\lim_{x \to 2} \left[ \frac{1}{x - 2} - \frac{2\left( 2x - 3 \right)}{x^3 - 3 x^2 + 2x} \right]\] 


\[\lim_{x \to a} \frac{\left( x + 2 \right)^{3/2} - \left( a + 2 \right)^{3/2}}{x -  a}\]


\[\lim_{x \to - 1/2} \frac{8 x^3 + 1}{2x + 1}\]


\[\lim_{x \to a} \frac{x^{2/3} - a^{2/3}}{x^{3/4} - a^{3/4}}\] 


\[\lim_{x \to \infty} \sqrt{x^2 + cx - x}\] 


\[\lim_{x \to \infty} \frac{3 x^{- 1} + 4 x^{- 2}}{5 x^{- 1} + 6 x^{- 2}}\]


\[\lim_{n \to \infty} \left[ \frac{1^3 + 2^3 + . . . n^3}{\left( n - 1 \right)^4} \right]\] 


Show that \[\lim_{x \to \infty} \left( \sqrt{x^2 + x + 1} - x \right) \neq \lim_{x \to \infty} \left( \sqrt{x^2 + 1} - x \right)\] 


\[\lim_{x \to 0} \frac{\sin x^0}{x}\] 


\[\lim_{x \to 0} \frac{\sin 5x}{\tan 3x}\] 


\[\lim_{x \to 0} \frac{\cos ax - \cos bx}{\cos cx - \cos dx}\] 


\[\lim_{x \to 0} \frac{x \cos x + 2 \sin x}{x^2 + \tan x}\] 


\[\lim_{x \to 0} \frac{5 x \cos x + 3 \sin x}{3 x^2 + \tan x}\] 


\[\lim_{x \to 0} \frac{2 \sin x^\circ - \sin 2 x^\circ}{x^3}\] 


\[\lim_{x \to 0} \frac{3 \sin x - \sin 3x}{x^3}\]


\[\lim_{x \to 0} \left( cosec x - \cot x \right)\]


Evaluate the following limits: 

\[\lim_{x \to 0} \frac{\cos ax - \cos bx}{\cos cx - 1}\] 


\[\lim_{x \to \frac{\pi}{2}} \frac{\sin 2x}{\cos x}\] 


\[\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{\cos x} - \sqrt{\sin x}}{x - \frac{\pi}{4}}\] 


\[\lim_{x \to 1} \frac{1 + \cos \pi x}{\left( 1 - x \right)^2}\] 


\[\lim_{x \to \pi} \frac{\sqrt{2 + \cos x} - 1}{\left( \pi - x \right)^2}\]


\[\lim_{x \to \frac{\pi}{4}} \frac{{cosec}^2 x - 2}{\cot x - 1}\]


\[\lim_{x \to \frac{\pi}{6}} \frac{\cot^2 x - 3}{cosec x - 2}\]


\[\lim_{x \to \frac{\pi}{4}} \frac{2 - {cosec}^2 x}{1 - \cot x}\] 


\[\lim_{x \to 0} \left( \cos x + a \sin bx \right)^{1/x}\]


\[\lim_{x \to a} \frac{x^n - a^n}{x - a}\]  is equal at 


\[\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{x}\] is equal to 


\[\lim_{x \to \infty} \frac{\left| x \right|}{x}\]  is equal to 


\[\lim_{x \to 0} \frac{\left| \sin x \right|}{x}\]


Evaluate the following limits: `lim_(x -> 2)[(x^(-3) - 2^(-3))/(x - 2)]`


Evaluate the following limit:

`lim_(x -> 5)[(x^3 - 125)/(x^5 - 3125)]`


Evaluate the following limits: `lim_(x -> 5)[(x^3 - 125)/(x^2 - 25)]`


Which of the following function is not continuous at x = 0?


Evaluate: `lim_(x -> 1) ((1 + x)^6 - 1)/((1 + x)^2 - 1)`


If f(x) = `{{:(1 if x  "is rational"),(-1 if x  "is rational"):}` is continuous on ______.


Evaluate the following limit:

`lim _ (x -> 5) [(x^3 - 125) / (x^5 - 3125)]`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×