मराठी

Lim X → a X N − a N X − a is Equal at

Advertisements
Advertisements

प्रश्न

\[\lim_{x \to a} \frac{x^n - a^n}{x - a}\]  is equal at 

पर्याय

  • nan 

  • nan−1 

  • na 

  •  1

     
MCQ
Advertisements

उत्तर

nan−1 

\[\lim_{x \to a} \frac{x^n - a^n}{x - a}\]
\[ = \lim_{x \to a^+} \frac{x^n - a^n}{x - a} \left[ \because f\left( x \right) exists, \lim_{x \to a} f\left( x \right) = \lim_{x \to a^+} f\left( x \right) \right]\]
\[ = \lim_{h \to 0} \frac{\left( a + h \right)^n - a^n}{a + h - a}\]
\[ = \lim_{h \to 0} a^n \frac{\left[ \left( 1 + \frac{h}{a} \right)^n - 1 \right]}{h}\]
\[ = a^n \lim_{h \to 0} \left[ 1 + n \cdot \frac{h}{a} + \frac{n\left( n - 1 \right)}{2!}\frac{h^2}{a^2} . . . + . . . - 1 \right]\]
\[ = a^n \lim_{h \to 0} \left[ \frac{n}{a} + \frac{h\left( h - 1 \right)}{2!} \frac{h}{a^2} + . . . \right]\]
\[ = a^n \frac{n}{a}\]
\[ = n a^{n - 1}\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 29: Limits - Exercise 29.13 [पृष्ठ ७८]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 11
पाठ 29 Limits
Exercise 29.13 | Q 11 | पृष्ठ ७८

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

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


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


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


\[\lim_{x \to 5} \frac{x^2 - 9x + 20}{x^2 - 6x + 5}\] 


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


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


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


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


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


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


\[\lim_{x \to - \infty} \left( \sqrt{x^2 - 8x} + x \right)\] 


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


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


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


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


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


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


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


Evaluate the following limits: 

\[\lim_{x \to 0} \frac{2\sin x - \sin2x}{x^3}\] 

 


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


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


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


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


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


\[\lim_{x \to 0^+} \left\{ 1 + \tan^2 \sqrt{x} \right\}^{1/2x}\]


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


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


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


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


If \[\lim_{x \to 1} \frac{x + x^2 + x^3 + . . . + x^n - n}{x - 1} = 5050\] then n equal


The value of \[\lim_{x \to \infty} \frac{\sqrt{1 + x^4} + \left( 1 + x^2 \right)}{x^2}\]  is


The value of \[\lim_{n \to \infty} \left\{ \frac{1 + 2 + 3 + . . . + n}{n + 2} - \frac{n}{2} \right\}\] 


Evaluate the following limit:

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


Evaluate the following limits: `lim_(y -> 1) [(2y - 2)/(root(3)(7 + y) - 2)]`


Evaluate the following Limits: `lim_(x -> "a") ((x + 2)^(5/3) - ("a" + 2)^(5/3))/(x - "a")`


Evaluate `lim_(h -> 0) ((a + h)^2 sin (a + h) - a^2 sina)/h`


Evaluate the following limit :

`lim_(x->5)[(x^3-125)/(x^5-3125)]`


Evaluate the following limit :

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


Evaluate the Following limit:

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


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×