English

Lim X → π / 2 2 − Cos X − 1 X ( X − π 2 )

Advertisements
Advertisements

Question

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

Advertisements

Solution

\[\lim_{x \to \frac{\pi}{2}} \left[ \frac{2^{- \cos x} - 1}{x\left( x - \frac{\pi}{2} \right)} \right]\]
\[ = \lim_{x \to \frac{\pi}{2}} \left[ \frac{2^{- \sin \left( \frac{\pi}{2} - x \right)} - 1}{x\left( x - \frac{\pi}{2} \right)} \right] \left\{ \because \cos x = \sin \left( \frac{\pi}{2} - x \right) \right\}\]
 = `\lim_{x \to \frac{\pi}{2}} \left[ \frac{2^\sin \left( x - \frac{\pi}{2} \right) - 1}{\left( x - \frac{\pi}{2} \right) \times x} \right]`
 =` \lim_{x \to \frac{\pi}{2}} \left[ \frac{2^\sin \left( x - \frac{\pi}{2} \right) - 1}{\sin \left( x - \frac{\pi}{2} \right)} \times \frac{\sin \left( x - \frac{\pi}{2} \right)}{\left( x - \frac{\pi}{2} \right)} \times \frac{1}{x} \right]`
\[ = \log_e 2 \times 1 \times \frac{1}{\frac{\pi}{2}}\]
\[ = \frac{2}{\pi} \log_e 2\]

 

shaalaa.com
  Is there an error in this question or solution?
Chapter 29: Limits - Exercise 29.1 [Page 72]

APPEARS IN

R.D. Sharma Mathematics [English] Class 11
Chapter 29 Limits
Exercise 29.1 | Q 43 | Page 72

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

Find `lim_(x -> 1)` f(x), where `f(x) = {(x^2 -1, x <= 1), (-x^2 -1, x > 1):}`


Evaluate `lim_(x -> 0) f(x)` where `f(x) = { (|x|/x, x != 0),(0, x = 0):}`


Find `lim_(x -> 0)` f(x), where `f(x) = {(x/|x|, x != 0),(0, x = 0):}`


Let a1, a2,..., an be fixed real numbers and define a function f ( x) = ( x − a1 ) ( x − a2 )...( x − an ).

What is `lim_(x -> a_1) f(x)` ? For some a ≠ a1, a2, ..., an, compute `lim_(x -> a) f(x)`


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


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


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


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


\[\lim_{x \to 3} \frac{\sqrt{x + 3} - \sqrt{6}}{x^2 - 9}\] 


\[\lim_{x \to 7} \frac{4 - \sqrt{9 + x}}{1 - \sqrt{8 - x}}\] 


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


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


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


\[\lim_{x \to 0} \frac{\sqrt{1 + 3x} - \sqrt{1 - 3x}}{x}\]


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


\[\lim_{x \to \sqrt{10}} \frac{\sqrt{7 + 2x} - \left( \sqrt{5} + \sqrt{2} \right)}{x^2 - 10}\] 


\[\lim_{x \to \sqrt{6}} \frac{\sqrt{5 + 2x} - \left( \sqrt{3} + \sqrt{2} \right)}{x^2 - 6}\] 

 


\[\lim_{x \to 0} \frac{a^x + b^x - 2}{x}\]


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


\[\lim_{x \to a} \frac{\log x - \log a}{x - a}\] 


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


\[\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{\log \left( 1 + x \right)}\] 


`\lim_{x \to \pi/2} \frac{a^\cot x - a^\cos x}{\cot x - \cos x}`


\[\lim_{x \to 5} \frac{e^x - e^5}{x - 5}\]


`\lim_{x \to \pi/2} \frac{e^\cos x - 1}{\cos x}`


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


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


`\lim_{x \to 0} \frac{e^\tan x - 1}{\tan x}`


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


\[\lim_{x \to a} \left\{ \frac{\sin x}{\sin a} \right\}^\frac{1}{x - a}\]


Write the value of \[\lim_{n \to \infty} \frac{1 + 2 + 3 + . . . + n}{n^2} .\]


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×