English

Evaluate: limx→2x2-43x-2-x+2 - Mathematics

Advertisements
Advertisements

Question

Evaluate: `lim_(x -> 2) (x^2 - 4)/(sqrt(3x - 2) - sqrt(x + 2))`

Sum
Advertisements

Solution

Given that `lim_(x -> 2) (x^2 - 4)/(sqrt(3x - 2) - sqrt(x + 2))`

Rationalizing the denominator, we get

= `lim_(x -> 2) ((x - 2)(x + 2) [sqrt(3x - 2) + sqrt(x + 2)])/([sqrt(3x - 2) - sqrt(x + 2)][sqrt(3x - 2) + sqrt(x + 2)])`

= `lim_(x -> 2) ((x - 2)(x + 2)[sqrt(3x - 2) + sqrt(x + 2)])/(3x - 2 - x - 2)`

= `lim_(x -> 2) ((x - 2)(x + 2)[sqrt((3x - 2)) + sqrt(x + 2)])/(2x - 4)`

= `lim_(x -> 2) ((x - 2)(x + 2) [sqrt((3x - 2)) + sqrt(x + 2)])/(2(x - 2))`

= `lim_(x -> 2) ((x + 2)[sqrt(3x - 2) + sqrt(x + 2)])/2` 

Taking limits, we have

`= ((2 + 2)[sqrt(6 - 2) + sqrt(2 + 2)])/2`

= `(4[2 + 2])/2`

= `(4 xx 4)/2`

= 8

shaalaa.com
  Is there an error in this question or solution?
Chapter 13: Limits and Derivatives - Exercise [Page 240]

APPEARS IN

NCERT Exemplar Mathematics [English] Class 11
Chapter 13 Limits and Derivatives
Exercise | Q 8 | Page 240

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

Find `lim_(x -> 0)` f(x), where `f(x) = {(x/|x|, x != 0),(0, x = 0):}`


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


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


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


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


\[\lim_{x \to 2} \frac{\sqrt{x^2 + 1} - \sqrt{5}}{x - 2}\] 


\[\lim_{x \to 7} \frac{4 - \sqrt{9 + x}}{1 - \sqrt{8 - x}}\] 


\[\lim_{x \to 1} \frac{\sqrt{5x - 4} - \sqrt{x}}{x^3 - 1}\] 


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


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


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


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


\[\lim_{x \to 1} \frac{\left( 2x - 3 \right) \left( \sqrt{x} - 1 \right)}{3 x^2 + 3x - 6}\]


\[\lim_{x \to 0} \frac{\sqrt{1 + x^2} - \sqrt{1 + x}}{\sqrt{1 + x^3} - \sqrt{1 + x}}\] 


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


\[\lim_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h}, x \neq 0\] 


\[\lim_{x \to \sqrt{6}} \frac{\sqrt{5 + 2x} - \left( \sqrt{3} + \sqrt{2} \right)}{x^2 - 6}\] 

 


\[\lim_{x \to 0} \frac{9^x - 2 . 6^x + 4^x}{x^2}\] 


\[\lim_{x \to 0} \frac{8^x - 4^x - 2^x + 1}{x^2}\]


\[\lim_{x \to 0} \frac{a^{mx} - b^{nx}}{x}\] 


\[\lim_{x \to 0} \frac{a^{mx} - b^{nx}}{\sin kx}\]


\[\lim_{x \to 0} \frac{a^x + b^ x - c^x - d^x}{x}\]


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


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


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


\[\lim_{x \to a} \frac{\log x - \log a}{x - a}\] 


\[\lim_{x \to 0} \frac{\log \left( 2 + x \right) + \log 0 . 5}{x}\]


\[\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{\log \left( 1 + x \right)}\] 


`\lim_{x \to \pi/2} \frac{a^\cot x - a^\cos x}{\cot x - \cos x}`


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


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


`\lim_{x \to 0} \frac{e^\tan x - 1}{x}`


\[\lim_{x \to a} \left\{ \frac{\sin x}{\sin a} \right\}^\frac{1}{x - a}\]


\[\lim_{x \to 0} \left\{ \frac{e^x + e^{- x} - 2}{x^2} \right\}^{1/ x^2}\]


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×