English

Lim X → 1 √ 5 X − 4 − √ X X 2 − 1

Advertisements
Advertisements

Question

\[\lim_{x \to 1} \frac{\sqrt{5x - 4} - \sqrt{x}}{x^2 - 1}\] 

Advertisements

Solution

\[\lim_{x \to 1} \left[ \frac{\sqrt{5x - 4} - \sqrt{x}}{x^2 - 1} \right]\] It is of the form \[\frac{0}{0}\] 

Rationalising the numerator: 

\[\lim_{x \to 1} \left[ \frac{\left( \sqrt{5x - 4} - \sqrt{x} \right)\left( \sqrt{5x - 4} + \sqrt{x} \right)}{\left( \sqrt{5x - 4} + \sqrt{x} \right)\left( x^2 - 1 \right)} \right]\] 

= \[\lim_{x \to 1} \left[ \frac{5x - 4 - x}{\left( \sqrt{5x - 4} + \sqrt{x} \right)\left( x - 1 \right)\left( x + 1 \right)} \right]\] 

=  \[\lim_{x \to 1} \left[ \frac{4\left( x - 1 \right)}{\left( \sqrt{5x - 4} + \sqrt{x} \right)\left( x - 1 \right)\left( x + 1 \right)} \right]\] 

= \[\frac{4}{\left( \sqrt{5 - 4} + \sqrt{1} \right)\left( 1 + 1 \right)}\] 

=  \[\frac{4}{2 \times 2}\] 

= 1 

shaalaa.com
  Is there an error in this question or solution?
Chapter 29: Limits - Exercise 29.4 [Page 28]

APPEARS IN

R.D. Sharma Mathematics [English] Class 11
Chapter 29 Limits
Exercise 29.4 | Q 11 | Page 28

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

Let a1, a2,..., an be fixed real numbers and define a function f ( x) = ( x − a1 ) ( x − a2 )...( x − an ).

What is `lim_(x -> a_1) f(x)` ? For some a ≠ a1, a2, ..., an, compute `lim_(x -> a) f(x)`


If f(x) = `{(|x| +  1,x < 0), (0, x = 0),(|x| -1, x > 0):}`

For what value (s) of a does `lim_(x -> a)`  f(x) exists?


If the function f(x) satisfies `lim_(x -> 1) (f(x) - 2)/(x^2 - 1) = pi`, evaluate `lim_(x -> 1) f(x)`.


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


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


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


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


\[\lim_{x \to 5} \frac{x - 5}{\sqrt{6x - 5} - \sqrt{4x + 5}}\] 


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


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


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


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

 


\[\lim_{x \to 0} \frac{\log \left( 1 + x \right)}{3^x - 1}\]


\[\lim_{x \to 0} \frac{a^{mx} - 1}{b^{nx} - 1}, n \neq 0\]


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


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


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


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


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


\[\lim_{x \to 0} \frac{\log \left| 1 + x^3 \right|}{\sin^3 x}\] 

 


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


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


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


\[\lim_{x \to 0} \frac{e^{bx} - e^{ax}}{x} \text{ where } 0 < a < b\] 


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


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


\[\lim_{x \to 0} \frac{x\left( e^x - 1 \right)}{1 - \cos x}\]


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


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


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


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


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


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


Let f(x) be a polynomial of degree 4 having extreme values at x = 1 and x = 2. If `lim_(x rightarrow 0) ((f(x))/x^2 + 1)` = 3 then f(–1) is equal to ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×