English

Lim X → 1 / 4 4 X − 1 2 √ X − 1 - Mathematics

Advertisements
Advertisements

Question

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

Advertisements

Solution

\[\lim_{x \to 1/4} \left[ \frac{4x - 1}{2\sqrt{x} - 1} \right]\]
\[\text{ It is of the form } \frac{0}{0} . \]
\[ \lim_{x \to 1/4} \left[ \frac{\left( 2\sqrt{x} \right)^2 - 1^2}{2\sqrt{x} - 1} \right]\]
\[ = \lim_{x \to 1/4} \left[ \frac{\left( 2\sqrt{x} - 1 \right)\left( 2\sqrt{x} + 1 \right)}{\left( 2\sqrt{x} - 1 \right)} \right]\]
\[ = 2\sqrt{\frac{1}{4}} + 1\]
\[ = 2 \times \frac{1}{2} + 1\]
\[ = 2\]

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

APPEARS IN

RD Sharma Mathematics [English] Class 11
Chapter 29 Limits
Exercise 29.3 | Q 18 | Page 23

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

\[\lim_{x \to a} \frac{\sqrt{x} + \sqrt{a}}{x + a}\] 


\[\lim_{x \to 0} 9\] 


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


\[\lim_{x \to 5} \frac{x^3 - 125}{x^2 - 7x + 10}\] 


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


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


\[\lim_{x \to a} \frac{\left( x + 2 \right)^{5/2} - \left( a + 2 \right)^{5/2}}{x - a}\] 


\[\lim_{x \to 4} \frac{x^3 - 64}{x^2 - 16}\] 


If \[\lim_{x \to a} \frac{x^9 - a^9}{x - a} = 9,\] find all possible values of a


\[\lim_{x \to \infty} \sqrt{x^2 + 7x - x}\] 


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


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


\[\lim_\theta \to 0 \frac{\sin 3\theta}{\tan 2\theta}\] 


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


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


\[\lim_{x \to 0} \frac{ax + x \cos x}{b \sin x}\]


\[\lim_{x \to 0} \frac{1 - \cos 5x}{1 - \cos 6x}\]


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


Evaluate the following limits: 

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


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


\[\lim_{x \to \frac{\pi}{3}} \frac{\sqrt{3} - \tan x}{\pi - 3x}\]


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


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


\[\lim_{x \to \frac{\pi}{2}} \left( \frac{\pi}{2} - x \right) \tan x\]


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


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


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


Write the value of \[\lim_{x \to 0^-} \left[ x \right] .\]


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


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


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


\[\lim_{n \to \infty} \left\{ \frac{1}{1 . 3} + \frac{1}{3 . 5} + \frac{1}{5 . 7} + . . . + \frac{1}{\left( 2n + 1 \right) \left( 2n + 3 \right)} \right\}\]is equal to


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


Evaluate the following limits: if `lim_(x -> 5)[(x^"k" - 5^"k")/(x - 5)]` = 500, find all possible values of k.


Let f(x) = `{{:(3^(1/x);   x < 0","                "then at"  x = 0),(lambda[x];   x ≥ 0","   lambda ∈ "R"):}`

`1/(ax^2 + bx + c)`


Evaluate the following limits: `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×