मराठी

The Value of Lim X → 0 √ a 2 − a X + X 2 − √ a 2 + a X + X 2 √ a + X − √ a − X

Advertisements
Advertisements

प्रश्न

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

पर्याय

  • \[\sqrt{a}\] 

  • \[- \sqrt{a}\]

MCQ
Advertisements

उत्तर

\[- \sqrt{a}\] 

\[\lim_{x \to 0} \frac{\sqrt{a^2 - ax + x^2} - \sqrt{a^2 + ax + x^2}}{\sqrt{a + x} - \sqrt{a - x}}\]
\[\text{ Multiplying } \left( \sqrt{a^2 - ax + x^2} + \sqrt{a^2 + ax + x^2} \right) and \left( \sqrt{a + x} + \sqrt{a - x} \right) in N^r and D^r : \]
\[ \lim_{x \to 0} \frac{\left( a^2 + x^2 - ax - a^2 - ax - x^2 \right) \left( \sqrt{a + x} + \sqrt{a - x} \right)}{\left( a + x - a + x \right) \left( \sqrt{x^2 - ax + x^2} + \sqrt{a^2 + ax + x^2} \right)}\]
\[ \lim_{x \to 0} \frac{- 2ax \left( \sqrt{a + x} + \sqrt{a - x} \right)}{2x \left( \sqrt{a^2 + ax + x^2} + \sqrt{a^2 + ax + x^2} \right)}\]
\[ = \frac{- a\left( 2\sqrt{a} \right)}{2a} = - \sqrt{a}\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 29: Limits - Exercise 29.13 [पृष्ठ ८०]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 11
पाठ 29 Limits
Exercise 29.13 | Q 31 | पृष्ठ ८०

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Show that \[\lim_{x \to 0} \frac{x}{\left| x \right|}\] does not exist.


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


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


\[\lim_{x \to - 1/2} \frac{8 x^3 + 1}{2x + 1}\] 


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


\[\lim_{x \to 1} \frac{x^3 + 3 x^2 - 6x + 2}{x^3 + 3 x^2 - 3x - 1}\]


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


\[\lim_{x \to - 1/2} \frac{8 x^3 + 1}{2x + 1}\]


\[\lim_{x \to 27} \frac{\left( x^{1/3} + 3 \right) \left( x^{1/3} - 3 \right)}{x - 27}\] 


\[\lim_{x \to 1} \frac{x^{15} - 1}{x^{10} - 1}\] 


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


\[\lim_{x \to 0} \frac{3 \sin x - 4 \sin^3 x}{x}\] 


\[\lim_{x \to 0} \frac{7x \cos x - 3 \sin x}{4x + \tan x}\] 


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


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


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


\[\lim_{x \to 0} \frac{1 - \cos 2x}{3 \tan^2 x}\] 


Evaluate the following limit: 

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


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


\[\lim_{x \to \frac{\pi}{8}} \frac{\cot 4x - \cos 4x}{\left( \pi - 8x \right)^3}\] 


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


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


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

 

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


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


Write the value of \[\lim_{x \to 0} \frac{\sin x^\circ}{x} .\]


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


\[\lim_{h \to 0} \left\{ \frac{1}{h\sqrt[3]{8 + h}} - \frac{1}{2h} \right\} =\]


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


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


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


\[\lim_{x \to \pi/4} \frac{4\sqrt{2} - \left( \cos x + \sin x \right)^5}{1 - \sin 2x}\] is equal to 


\[\lim_{x \to \infty} a^x \sin \left( \frac{b}{a^x} \right), a, b > 1\] is equal to 


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


Let `f(x) = {{:((k cos x)/(pi - 2x)",", "when"  x ≠ pi/2),(3",", x = pi/2  "and if"  f(x) = f(pi/2)):}` find the value of k.


If `f(x) = {{:(x + 2",",  x ≤ - 1),(cx^2",", x > -1):}`, find 'c' if `lim_(x -> -1) f(x)` exists


Evaluate the following limits: `lim_(x ->3) [sqrt(x + 6)/x]`


Evaluate the Following limit:

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


Evaluate the following limit :

`lim_(x->7)[[(root3(x)- root3(7))(root3(x) + root3(7)))/(x-7)]`


Evaluate the following limits: `lim_(x -> 5)[(x^3 - 125)/(x^5 - 3125)]`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×