English

Lim X → √ 3 X 4 − 9 X 2 + 4 √ 3 X − 15

Advertisements
Advertisements

Question

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

Advertisements

Solution

\[\lim_{x \to \sqrt{3}} \left[ \frac{x^4 - 9}{x^2 + 4\sqrt{3}x - 15} \right]\]
\[\text{ It is of the form } \frac{0}{0} . \]
\[ \lim_{x \to \sqrt{3}} \left[ \frac{\left( x^2 \right)^2 - \left( 3 \right)^2}{x^2 + 5\sqrt{3}x - \sqrt{3}x - 15} \right]\]
\[ = \lim_{x \to \sqrt{3}} \left[ \frac{\left( x^2 - 3 \right)\left( x^2 + 3 \right)}{x\left( x + 5\sqrt{3} \right) - \sqrt{3}\left( x + 5\sqrt{3} \right)} \right]\]
\[ = \lim_{x \to \sqrt{3}} \left[ \frac{\left\{ x^2 - \left( \sqrt{3} \right)^2 \right\}\left( x^2 + 3 \right)}{\left( x - \sqrt{3} \right)\left( x + 5\sqrt{3} \right)} \right]\]
\[ = \lim_{x \to \sqrt{3}} \left[ \frac{\left( x - \sqrt{3} \right)\left( x + \sqrt{3} \right)\left( x^2 + 3 \right)}{\left( x - \sqrt{3} \right)\left( x + 5\sqrt{3} \right)} \right]\]
\[ = \frac{\left( \sqrt{3} + \sqrt{3} \right)\left( 3 + 3 \right)}{\sqrt{3} + 5\sqrt{3}}\]
\[ = \frac{2\sqrt{3} \times 6}{6\sqrt{3}}\]
\[ = 2\]

 

shaalaa.com
  Is there an error in this question or solution?
Chapter 29: Limits - Exercise 29.3 [Page 23]

APPEARS IN

R.D. Sharma Mathematics [English] Class 11
Chapter 29 Limits
Exercise 29.3 | Q 13 | Page 23

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

\[\lim_{x \to 0} 9\] 


\[\lim_{x \to 4} \frac{x^2 - 7x + 12}{x^2 - 3x - 4}\] 


\[\lim_{x \to 1/4} \frac{4x - 1}{2\sqrt{x} - 1}\] 


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


\[\lim_{x \to 2} \frac{x^3 + 3 x^2 - 9x - 2}{x^3 - x - 6}\] 


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


\[\lim_{x \to a} \frac{x^{5/7} - a^{5/7}}{x^{2/7} - a^{2/7}}\] 


`lim_(x->∞) [x{sqrt(x^2+1) - sqrt(x^2-1)}]`


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


\[\lim_{x \to \infty} \left[ \frac{x^4 + 7 x^3 + 46x + a}{x^4 + 6} \right]\] where a is a non-zero real number. 


\[\lim_{x \to - \infty} \left( \sqrt{4 x^2 - 7x} + 2x \right)\] 


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


\[\lim_{x \to 0} \frac{\tan 8x}{\sin 2x}\] 


\[\lim_{x \to 0} \frac{\cos 3x - \cos 7x}{x^2}\] 


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


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


\[\lim_{x \to 0} \frac{\sqrt{2} - \sqrt{1 + \cos x}}{x^2}\] 


\[\lim_{x \to 0} \frac{x^2 + 1 - \cos x}{x \sin x}\] 


\[\lim_{x \to 0} \frac{\sin 2x \left( \cos 3x - \cos x \right)}{x^3}\] 


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


\[\lim_{x \to 0} \frac{x^3 \cot x}{1 - \cos x}\] 


\[\lim_{x \to 0} \frac{\cos 2x - 1}{\cos x - 1}\] 


\[\lim_{x \to 0} \frac{\sin 3x + 7x}{4x + \sin 2x}\]


\[\lim_{x \to 0} \frac{5x + 4 \sin 3x}{4 \sin 2x + 7x}\]


Evaluate the following limit: 

\[\lim_{x \to 0} \frac{\sin\left( \alpha + \beta \right)x + \sin\left( \alpha - \beta \right)x + \sin2\alpha x}{\cos^2 \beta x - \cos^2 \alpha x}\]


Evaluate the following limit:

\[\lim_{x \to \frac{\pi}{3}} \frac{\sqrt{1 - \cos6x}}{\sqrt{2}\left( \frac{\pi}{3} - x \right)}\]


\[\lim_{x \to \frac{\pi}{2}} \frac{\cot x}{\frac{\pi}{2} - x}\]


\[\lim_{x \to \frac{\pi}{4}} \frac{1 - \sin 2x}{1 + \cos 4x}\] 


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


Evaluate the following limit:

\[\lim_{x \to \pi} \frac{1 - \sin\frac{x}{2}}{\cos\frac{x}{2}\left( \cos\frac{x}{4} - \sin\frac{x}{4} \right)}\]

 


\[\lim_{x \to 0} \frac{\sin 2x}{e^x - 1}\] 


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


Write the value of \[\lim_{x \to \infty} \frac{\sin x}{x} .\] 


Write the value of \[\lim_{x \to 1^-} x - \left[ x \right] .\] 


Write the value of \[\lim_{x \to \pi} \frac{\sin x}{x - \pi} .\]


\[\lim_{x \to \infty} \frac{\sin x}{x}\] equals 


The value of \[\lim_{x \to \infty} \frac{\sqrt{1 + x^4} + \left( 1 + x^2 \right)}{x^2}\]  is


\[\lim_{n \to \infty} \frac{n!}{\left( n + 1 \right)! + n!}\]  is equal to


Evaluate the following limit :

`lim_(x->3)[sqrt(x+6)/x]`


Evaluate the following limit.

`lim_(x->3)[sqrt(x + 6)/x]`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×