मराठी

Lim X → 0 Tan 8 X Sin 2 X

Advertisements
Advertisements

प्रश्न

\[\lim_{x \to 0} \frac{\tan 8x}{\sin 2x}\] 

Advertisements

उत्तर

\[\lim_{x \to 0} \left[ \frac{\tan 8x}{\sin 2x} \right]\] 

\[\Rightarrow \lim_{x \to 0} \left[ \frac{\tan 8x}{8x} \times \frac{8x}{\frac{\sin 2x}{2x} \times 2x} \right] \left[ \because \lim_{x \to 0} \frac{\tan 8x}{8x} = 1, \lim_{x \to 0} \frac{\sin 2x}{2x} = 1 \right]\]
\[ \Rightarrow 4\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 29: Limits - Exercise 29.7 [पृष्ठ ५०]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 11
पाठ 29 Limits
Exercise 29.7 | Q 6 | पृष्ठ ५०

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?


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


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


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


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


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


\[\lim_{x \to \infty} \frac{3 x^3 - 4 x^2 + 6x - 1}{2 x^3 + x^2 - 5x + 7}\] 


\[\lim_{x \to \infty} \sqrt{x^2 + 7x - x}\] 


\[f\left( x \right) = \frac{a x^2 + b}{x^2 + 1}, \lim_{x \to 0} f\left( x \right) = 1\] and \[\lim_{x \to \infty} f\left( x \right) = 1,\]then prove that f(−2) = f(2) = 1


\[\lim_{x \to 0} \frac{\cos ax - \cos bx}{\cos cx - \cos dx}\] 


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


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


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


\[\lim_{x \to 0} \frac{x^3 \cot x}{1 - \cos x}\] 


\[\lim_{x \to 0} \frac{cosec x - \cot x}{x}\]


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


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


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


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


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


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


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


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


Evaluate the following limit:

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

 


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


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


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


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


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


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


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


\[\lim 2_{h \to 0} \left\{ \frac{\sqrt{3} \sin \left( \pi/6 + h \right) - \cos \left( \pi/6 + h \right)}{\sqrt{3} h \left( \sqrt{3} \cos h - \sin h \right)} \right\}\] 


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


Evaluate the following limit:

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


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


Number of values of x where the function

f(x) = `{{:((tanxlog(x - 2))/(x^2 - 4x + 3); x∈(2, 4) - {3, π}),(1/6tanx; x = 3","  π):}`

is discontinuous, is ______.


Evaluate the following limit:

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


Evaluate the following limit:

`\underset{x->3}{lim}[sqrt(x +6)/(x)]`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×