मराठी

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

Advertisements
Advertisements

प्रश्न

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

Advertisements

उत्तर

\[\lim_{x \to \frac{\pi}{2}} \left[ \frac{2^{- \cos x} - 1}{x\left( x - \frac{\pi}{2} \right)} \right]\]
\[ = \lim_{x \to \frac{\pi}{2}} \left[ \frac{2^{- \sin \left( \frac{\pi}{2} - x \right)} - 1}{x\left( x - \frac{\pi}{2} \right)} \right] \left\{ \because \cos x = \sin \left( \frac{\pi}{2} - x \right) \right\}\]
 = `\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
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 29: Limits - Exercise 29.1 [पृष्ठ ७२]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 11
पाठ 29 Limits
Exercise 29.1 | Q 43 | पृष्ठ ७२

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Find `lim_(x -> 0)` f(x) and `lim_(x -> 1)` f(x) where f(x) = `{(2x + 3, x <= 0),(3(x+1), x > 0):}`


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


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


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


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


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


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


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


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


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


\[\lim_{x \to 1} \frac{\left( 2x - 3 \right) \left( \sqrt{x} - 1 \right)}{3 x^2 + 3x - 6}\]


\[\lim_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h}, x \neq 0\] 


\[\lim_{x \to 0} \frac{a^{mx} - 1}{b^{nx} - 1}, n \neq 0\]


\[\lim_{x \to 0} \frac{9^x - 2 . 6^x + 4^x}{x^2}\] 


\[\lim_{x \to 0} \frac{8^x - 4^x - 2^x + 1}{x^2}\]


\[\lim_{x \to 0} \frac{a^{mx} - b^{nx}}{x}\] 


\[\lim_{x \to 0} \frac{5^x + 3^x + 2^x - 3}{x}\]


\[\lim_{x \to 0} \frac{a^{mx} - b^{nx}}{\sin kx}\]


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


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


\[\lim_{x \to 0} \frac{\log \left( 2 + x \right) + \log 0 . 5}{x}\]


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


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


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


`\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 0} \frac{e^x - e^\sin x}{x - \sin x}`


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


\[\lim_{x \to \infty} \left\{ \frac{x^2 + 2x + 3}{2 x^2 + x + 5} \right\}^\frac{3x - 2}{3x + 2}\]


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


Write the value of \[\lim_{n \to \infty} \frac{n! + \left( n + 1 \right)!}{\left( n + 1 \right)! + \left( n + 2 \right)!} .\]


Write the value of \[\lim_{x \to \pi/2} \frac{2x - \pi}{\cos x} .\] 


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


Evaluate: `lim_(x -> 2) (x^2 - 4)/(sqrt(3x - 2) - sqrt(x + 2))`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×