हिंदी

Lim X → a ( X + 2 ) 3 / 2 − ( a + 2 ) 3 / 2 X − a

Advertisements
Advertisements

प्रश्न

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

Advertisements

उत्तर

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

Let y = x + 2 and b = a + 2.

When x ​→ a and x + 2 ​→ a + 2. 

\[\Rightarrow\]y ​→ b 

\[\lim_{y \to b} \left[ \frac{y^\frac{3}{2} - b^\frac{3}{2}}{y - b} \right]\]
\[ = \frac{3}{2} \left( b \right)^\frac{3}{2} - 1 \]
\[ = \frac{3}{2} b^\frac{1}{2} \]
\[ = \frac{3}{2} \left( a + 2 \right)^\frac{1}{2}\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 29: Limits - Exercise 29.5 [पृष्ठ ३३]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 29 Limits
Exercise 29.5 | Q 2 | पृष्ठ ३३

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

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


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


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


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


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


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


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


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


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


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


\[\lim_{x \to \infty} \frac{3 x^3 - 4 x^2 + 6x - 1}{2 x^3 + x^2 - 5x + 7}\] 


\[\lim_{x \to \infty} \left[ \left\{ \sqrt{x + 1} - \sqrt{x} \right\} \sqrt{x + 2} \right]\] 


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


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


\[\lim_{x \to 0} \frac{\tan mx}{\tan nx}\] 


\[\lim_{x \to 0} \frac{\sin 2x \left( \cos 3x - \cos x \right)}{x^3}\] 


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


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


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


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


\[\lim_{x \to 0} \frac{1 - \cos 4x}{x^2}\] 


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


\[\lim_{x \to 0} \frac{cosec x - \cot x}{x}\]


Evaluate the following limit: 

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


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


\[\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{\cos x} - \sqrt{\sin x}}{x - \frac{\pi}{4}}\] 


\[\lim_{x \to \frac{\pi}{4}} \frac{f\left( x \right) - f\left( \frac{\pi}{4} \right)}{x - \frac{\pi}{4}},\]


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


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


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


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


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


\[\lim_{x \to 1} \left[ x - 1 \right]\] where [.] is the greatest integer function, is equal to 


Evaluate the following limit:

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


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.


Evaluate the following limit :

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


Evaluate the following limit :

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


Evaluate the following limit:

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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×