हिंदी

lim x → 3 x − 3 | x − 3 | , is equal to - Mathematics

Advertisements
Advertisements

प्रश्न

\[\lim_{x \to 3} \frac{x - 3}{\left| x - 3 \right|},\] is equal to

विकल्प

  •  1 

  • −1 

  •  0 

  • does not exist 

MCQ
Advertisements

उत्तर

does not exist 

\[\lim_{x \to 3} \frac{x - 3}{\left| x - 3 \right|}\]
\[\text{ LHL at } x = 3: \]
\[ \lim_{x \to 3^-} \frac{x - 3}{- \left( x - 3 \right)} \left[ \because \left| x - 3 \right| = - \left( x - 3 \right), \text{ when } x < 3 \right]\]
\[ \Rightarrow - 1\]
\[\text{ RHL at } x = 3: \]
\[ \lim_{x \to 3^+} \frac{x - 3}{x - 3} \left[ \because \left| x - 3 \right| = x - 3, \text{ when } x > 3 \right]\]
\[ = 1\]

LHL ≠ RHL
Therefore, limit does not exist. 

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 29: Limits - Exercise 29.13 [पृष्ठ ७८]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
अध्याय 29 Limits
Exercise 29.13 | Q 10 | पृष्ठ ७८

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

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


\[\lim_{x \to 2} \frac{x^3 - 8}{x^2 - 4}\] 


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


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


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


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


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


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


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


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


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


Evaluate: \[\lim_{n \to \infty} \frac{1^4 + 2^4 + 3^4 + . . . + n^4}{n^5} - \lim_{n \to \infty} \frac{1^3 + 2^3 + . . . + n^3}{n^5}\] 


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


Write the value of \[\lim_{x \to \pi} \frac{\sin x}{x - \pi} .\]


\[\lim_{x \to 0} \frac{\sin 2x}{x}\] 


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


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


\[\lim_{x \to \pi/3} \frac{\sin \left( \frac{\pi}{3} - x \right)}{2 \cos x - 1}\] is equal to 


Evaluate the Following limit:

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


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×