English

Lim X → 1 Sin π X X − 1 - Mathematics

Advertisements
Advertisements

Question

\[\lim_{x \to 1} \frac{\sin \pi x}{x - 1}\] 

Options

  • π 

  • π 

  • \[- \frac{1}{\pi}\] 

  • \[\frac{1}{\pi}\] 

MCQ

Solution

 −π 

\[\lim_{x \to 1} \frac{\sin \pi x}{x - 1}\]
\[ = \lim_{h \to 0} \frac{\sin \pi\left( 1 + h \right)}{\left( 1 + h \right) - 1}\]
\[ = \lim_{h \to 0} \frac{\sin \left( \pi + \pi h \right)}{h}\]
\[ = \lim_{h \to 0} \frac{- \sin \pi h}{h}\]
\[ = \lim_{h \to 0} - \left( \frac{\sin \pi h}{\pi h} \right)\pi\]
\[ = - \pi\] 

shaalaa.com
  Is there an error in this question or solution?
Chapter 29: Limits - Exercise 29.13 [Page 79]

APPEARS IN

RD Sharma Mathematics [English] Class 11
Chapter 29 Limits
Exercise 29.13 | Q 17 | Page 79

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

Show that \[\lim_{x \to 0} \frac{x}{\left| x \right|}\] does not exist.


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


\[\lim_{x \to 0} \frac{ax + b}{cx + d}, d \neq 0\]


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


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


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


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


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


\[\lim_{x \to \infty} \left[ \frac{x^4 + 7 x^3 + 46x + a}{x^4 + 6} \right]\] where a is a non-zero real number. 


\[\lim_{x \to - \infty} \left( \sqrt{4 x^2 - 7x} + 2x \right)\] 


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


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


\[\lim_{x \to 0} \frac{3 \sin x - 4 \sin^3 x}{x}\] 


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


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


\[\lim_\theta \to 0 \frac{\sin 4\theta}{\tan 3\theta}\] 


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


\[\lim_{x \to \frac{\pi}{4}} \frac{1 - \sin 2x}{1 + \cos 4x}\] 


\[\lim_{x \to \frac{\pi}{6}} \frac{\cot^2 x - 3}{cosec x - 2}\]


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


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


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


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


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

 

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


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


\[\lim_{x \to 0} \frac{\sin 2x}{x}\] 


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


\[\lim_{x \to \pi/4} \frac{\sqrt{2} \cos x - 1}{\cot x - 1}\] is equal to


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


\[\lim_{x \to 2} \frac{\sqrt{1 + \sqrt{2 + x} - \sqrt{3}}}{x - 2}\] is equal to 


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


\[\lim_{x \to 0} \frac{\left| \sin x \right|}{x}\]


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


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


Evaluate the following limits: `lim_(x -> 3) [sqrt(x + 6)/x]`


Evaluate the following limits: `lim_(x -> 5)[(x^3 - 125)/(x^5 - 3125)]`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×