हिंदी

Lim X → 3 X 2 − X − 6 X 3 − 3 X 2 + X − 3

Advertisements
Advertisements

प्रश्न

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

Advertisements

उत्तर

\[\lim_{x \to 3} \left[ \frac{x^2 - x - 6}{x^3 - 3 x^2 + x - 3} \right]\]
\[\text{ It is of the form } \frac{0}{0} . \]
\[ \lim_{x \to 3} \left[ \frac{x^2 - 3x + 2x - 6}{x^2 \left( x - 3 \right) + 1\left( x - 3 \right)} \right]\]
\[ = \lim_{x \to 3} \left[ \frac{x\left( x - 3 \right) + 2\left( x - 3 \right)}{\left( x^2 + 1 \right)\left( x - 3 \right)} \right]\]
\[ = \lim_{x \to 3} \left[ \frac{\left( x + 2 \right)\left( x - 3 \right)}{\left( x^2 + 1 \right)\left( x - 3 \right)} \right]\]
\[ = \frac{3 + 2}{3^2 + 1}\]
\[ = \frac{5}{10}\]
\[ = \frac{1}{2}\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 29: Limits - Exercise 29.3 [पृष्ठ २३]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 29 Limits
Exercise 29.3 | Q 28 | पृष्ठ २३

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

Find `lim_(x -> 5) f(x)`, where f(x)  = |x| - 5


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


\[\lim_{x \to 0} \frac{x^{2/3} - 9}{x - 27}\]


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


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


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


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


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


If \[\lim_{x \to a} \frac{x^3 - a^3}{x - a} = \lim_{x \to 1} \frac{x^4 - 1}{x - 1},\] find all possible values of a


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


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


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


\[\lim_{n \to \infty} \left[ \frac{\left( n + 2 \right)! + \left( n + 1 \right)!}{\left( n + 2 \right)! - \left( n + 1 \right)!} \right]\] 


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


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


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


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


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


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


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


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


Evaluate the following limit:

\[\lim_{x \to \pi} \frac{1 - \sin\frac{x}{2}}{\cos\frac{x}{2}\left( \cos\frac{x}{4} - \sin\frac{x}{4} \right)}\]

 


\[\lim_{x \to \frac{3\pi}{2}} \frac{1 + {cosec}^3 x}{\cot^2 x}\]


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


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


Write the value of \[\lim_{x \to 0} \frac{\sqrt{1 - \cos 2x}}{x} .\]


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


\[\lim_{x \to 3} \frac{\sum^n_{r = 1} x^r - \sum^n_{r = 1} 3^r}{x - 3}\]is real to


\[\lim_{x \to 0} \frac{8}{x^8}\left\{ 1 - \cos \frac{x^2}{2} - \cos \frac{x^2}{4} + \cos \frac{x^2}{2} \cos \frac{x^2}{4} \right\}\] is equal to 


The value of \[\lim_{x \to 0} \frac{1 - \cos x + 2 \sin x - \sin^3 x - x^2 + 3 x^4}{\tan^3 x - 6 \sin^2 x + x - 5 x^3}\] is 


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 


If `lim_(x -> 1) (x^4 - 1)/(x - 1) = lim_(x -> k) (x^3 - l^3)/(x^2 - k^2)`, then find the value of k.


Evaluate the Following limit: 

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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×