मराठी

Lim X → − 1 X 2 − X − 2 ( X 2 + X ) + Sin ( X + 1 )

Advertisements
Advertisements

प्रश्न

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

Advertisements

उत्तर

\[\lim_{x \to - 1} \frac{x^2 - x - 2}{x^2 + x + \sin \left( x + 1 \right)}\]
\[ = \lim_{x \to - 1} \frac{\left( x - 2 \right) \left( x + 1 \right)}{x\left( x + 1 \right) + \sin \left( x + 1 \right)}\]
\[Let y = x + 1\]
\[If x \to - 1, then y \to 0 . \]
\[ = \lim_{y \to 0} \frac{\left( y - 3 \right)y}{\left( y - 1 \right)y + \sin y}\]
\[\text{ Dividing the numerator and the denominator by } y:\]
\[ = \lim_{y \to 0} \frac{\left( y - 3 \right)}{\left( y - 1 \right) + \frac{\sin y}{y}}\]
\[ = \frac{0 - 3}{\left( 0 - 1 \right) + 1}\]
\[ \Rightarrow \frac{- 3}{0} = \infty\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 29: Limits - Exercise 29.8 [पृष्ठ ६२]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 11
पाठ 29 Limits
Exercise 29.8 | Q 27 | पृष्ठ ६२

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

\[\lim_{x \to 4} \frac{x^2 - 7x + 12}{x^2 - 3x - 4}\] 


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


\[\lim_{x \to 2} \frac{x^3 + 3 x^2 - 9x - 2}{x^3 - x - 6}\] 


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


Evaluate the following limit:

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


\[\lim_{x \to a} \frac{x^{5/7} - a^{5/7}}{x^{2/7} - a^{2/7}}\] 


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


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


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


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


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


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


\[\lim_{x \to 0} \frac{1 - \cos 2x}{\cos 2x - \cos 8x}\]


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


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


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


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


\[\lim_{x \to 0} \frac{5x + 4 \sin 3x}{4 \sin 2x + 7x}\]


Evaluate the following limit: 

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


\[\lim_{x \to a} \frac{\cos x - \cos a}{x - a}\] 


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


\[\lim_{x \to a} \frac{\sin \sqrt{x} - \sin \sqrt{a}}{x - a}\] 


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


\[\lim_{x \to \frac{\pi}{4}} \frac{{cosec}^2 x - 2}{\cot x - 1}\]


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


\[\lim_{x \to \infty} \frac{\sin x}{x} .\] 


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


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


If α is a repeated root of ax2 + bx + c = 0, then \[\lim_{x \to \alpha} \frac{\tan \left( a x^2 + bx + c \right)}{\left( x - \alpha \right)^2}\]


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


Evaluate the following limits: `lim_(x -> 0)[((1 - x)^8 - 1)/((1 - x)^2 - 1)]`


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


Evaluate the following limits: `lim_(x -> 5)[(x^3 - 125)/(x^2 - 25)]`


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


`lim_(x->3) (x^5 - 243)/(x^3 - 27)` = ?


Which of the following function is not continuous at x = 0?


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×