English

Lim X → 0 ( a + X ) 2 − a 2 X - Mathematics

Advertisements
Advertisements

Question

\[\lim_{x \to 0} \frac{\left( a + x \right)^2 - a^2}{x}\] 

Advertisements

Solution

\[\lim_{x \to 0} \left[ \frac{\left( a + x \right)^2 - a^2}{x} \right]\]
\[\text{ It is of the form } \frac{0}{0} . \]
\[ \lim_{x \to 0} \left[ \frac{a^2 + x^2 + 2ax - a^2}{x} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{x\left( x + 2a \right)}{x} \right]\]
\[ = \lim_{x \to 0} \left[ x + 2a \right]\]
\[ = 0 + 2a\]
\[ = 2a\]

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 20 | Page 23

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

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


\[\lim_{x \to 1} \frac{\sqrt{x + 8}}{\sqrt{x}}\] 


\[\lim_{x \to 2} \left( 3 - x \right)\] 


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


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


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


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


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


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


\[\lim_{x \to 0} \frac{\left( 1 + x \right)^6 - 1}{\left( 1 + x \right)^2 - 1}\] 


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


\[\lim_{x \to \infty} \frac{\sqrt{x^2 + a^2} - \sqrt{x^2 + b^2}}{\sqrt{x^2 + c^2} - \sqrt{x^2 + d^2}}\] 


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


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


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


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


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


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


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


Evaluate the following limit: 

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


If  \[\lim_{x \to 0} kx  cosec x = \lim_{x \to 0} x  cosec kx,\] 


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


\[\lim_{x \to \frac{\pi}{4}} \frac{1 - \tan x}{x - \frac{\pi}{4}}\] 


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


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


\[\lim_{n \to \infty} \frac{\sin \left( \frac{a}{2^n} \right)}{\sin \left( \frac{b}{2^n} \right)}\]


\[\lim_{x \to \pi} \frac{1 + \cos x}{\tan^2 x}\] 


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


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


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


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


\[\lim_{n \to \infty} \frac{1 - 2 + 3 - 4 + 5 - 6 + . . . . + \left( 2n - 1 \right) - 2n}{\sqrt{n^2 + 1} + \sqrt{n^2 - 1}}\] is equal to 


The value of \[\lim_{x \to \infty} \frac{\left( x + 1 \right)^{10} + \left( x + 2 \right)^{10} + . . . + \left( x + 100 \right)^{10}}{x^{10} + {10}^{10}}\] is 


Evaluate the following limit:

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


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


Evaluate the following limits: `lim_(x -> 0)[(root(3)(1 + x) - sqrt(1 + x))/x]`


if `lim_(x -> 2) (x^"n"- 2^"n")/(x - 2)` = 80 then find the value of n.


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


Evaluate the Following limit:

`lim_(x->7)[[(root[3][x] - root[3][7])(root[3][x] + root[3][7])] / (x - 7)]`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×