मराठी

If F ( X ) = { X Sin 1 X , X ≠ 0 0 , X = 0 , Then Lim X → 0 F ( X ) Equals

Advertisements
Advertisements

प्रश्न

If \[f\left( x \right) = \left\{ \begin{array}{l}x \sin \frac{1}{x}, & x \neq 0 \\ 0, & x = 0\end{array}, \right.\] then \[\lim_{x \to 0} f\left( x \right)\]  equals 

पर्याय

  •  1 

  •  0 

  •  −1 

  •  none of these 

MCQ
Advertisements

उत्तर

 0 

\[f\left( x \right) = \binom{x\sin\left( \frac{1}{x} \right), x \neq 0}{0, x = 0}\] 

LHL: 

\[\lim_{x \to 0^-} f\left( x \right)\]
\[ = \lim_{x \to 0^-} \left[ x\sin\left( \frac{1}{x} \right) \right]\]


Let x = 0 – h, where h → 0. 

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


= 0 × The oscillating number between –1 and 1
= 0
RHL 

\[\lim_{x \to 0^+} f\left( x \right)\] 

Let x = 0 + h, where h → 0.

\[= \lim_{h \to 0} \left[ h \times \sin\left( \frac{1}{h} \right) \right]\] 


= 0 × The oscillating number between –1 and 1
= 0
LHL = RHL = 0 

\[\therefore \lim_{x \to 0} f\left( x \right) = 0\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 29: Limits - Exercise 29.13 [पृष्ठ ७९]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 11
पाठ 29 Limits
Exercise 29.13 | Q 24 | पृष्ठ ७९

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Suppose f(x)  = `{(a+bx, x < 1),(4, x = 1),(b-ax, x > 1):}`  and if `lim_(x -> 1) f(x) = f(1)` what are possible values of a and b?


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


\[\lim_{x \to a} \frac{\sqrt{x} + \sqrt{a}}{x + a}\] 


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


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


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


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


Evaluate the following limit:

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


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


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


\[\lim_{x \to \infty} \frac{\sqrt{x^2 + a^2} - \sqrt{x^2 + b^2}}{\sqrt{x^2 + c^2} - \sqrt{x^2 + d^2}}\] 


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


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


\[\lim_{x \to 0} \frac{\sin 3x}{5x}\] 


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


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


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


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


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


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


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


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


Evaluate the following limit:

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

 


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


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


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


Write the value of \[\lim_{x \to 1^-} x - \left[ x \right] .\] 


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


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


\[\lim_{x \to a} \frac{x^n - a^n}{x - a}\]  is equal at 


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


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


\[\lim_{n \to \infty} \frac{1 - 2 + 3 - 4 + 5 - 6 + . . . . + \left( 2n - 1 \right) - 2n}{\sqrt{n^2 + 1} + \sqrt{n^2 - 1}}\] is equal to 


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


Evaluate the following limits: `lim_(x -> "a")[((z + 2)^(3/2) - ("a" + 2)^(3/2))/(z - "a")]`


Evaluate the Following limit:

`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×