हिंदी

Lim X → π 2 1 − Sin X ( π 2 − X ) 2 - Mathematics

Advertisements
Advertisements

प्रश्न

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

Advertisements

उत्तर

\[\lim_{x \to \frac{\pi}{2}} \frac{1 - \sin x}{\left( \frac{\pi}{2} - x \right)^2}\]
\[ = \lim_{h \to 0} \frac{1 - \sin \left( \frac{\pi}{2} - h \right)}{\left( \frac{\pi}{2} - \left( \frac{\pi}{2} - h \right) \right)^2}\]
\[ = \lim_{h \to 0} \frac{1 - \cos h}{h^2}\]
\[ = \lim_{h \to 0} \frac{2 \sin^2 \frac{h}{2}}{\frac{4 h^2}{4}} \left[ \because \lim_{h \to 0} \frac{\sin h}{h} = 1 \right]\]
\[ = \frac{1}{2} \lim_{h \to 0} \left( \frac{\sin \frac{h}{2}}{\frac{h}{2}} \right)^2 \]
\[ \Rightarrow \frac{1}{2}\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 29: Limits - Exercise 29.8 [पृष्ठ ६२]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
अध्याय 29 Limits
Exercise 29.8 | Q 7 | पृष्ठ ६२

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

\[\lim_{x \to 0} \frac{2 x^2 + 3x + 4}{x^2 + 3x + 2}\] 


\[\lim_{x \to 0} 9\] 


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


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


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


\[\lim_{x \to a} \frac{x^{2/7} - a^{2/7}}{x - a}\] 


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


`lim_(x->∞) [x{sqrt(x^2+1) - sqrt(x^2-1)}]`


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


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


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


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


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


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


Evaluate the following limit: 

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


\[\lim_{x \to \frac{\pi}{4}} \frac{f\left( x \right) - f\left( \frac{\pi}{4} \right)}{x - \frac{\pi}{4}},\]


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


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


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


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


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


If \[\lim_{x \to 1} \frac{x + x^2 + x^3 + . . . + x^n - n}{x - 1} = 5050\] then n equal


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


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


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


\[\lim_{x \to \infty} a^x \sin \left( \frac{b}{a^x} \right), a, b > 1\] is equal 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 \[\lim_{n \to \infty} \left\{ \frac{1 + 2 + 3 + . . . + n}{n + 2} - \frac{n}{2} \right\}\] 


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


Evaluate the following limit:

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


Evaluate the following limits: if `lim_(x -> 5)[(x^"k" - 5^"k")/(x - 5)]` = 500, find all possible values of k.


if `lim_(x -> 2) (x^"n"- 2^"n")/(x - 2)` = 80 then find the value of n.


Evaluate the following Limit:

`lim_(x -> 0) ((1 + x)^"n" - 1)/x`


Let f(x) = `{{:(3^(1/x);   x < 0","                "then at"  x = 0),(lambda[x];   x ≥ 0","   lambda ∈ "R"):}`

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


Evaluate the following limit :

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


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×