English

Lim X → a { Sin X Sin a } 1 X − a - Mathematics

Advertisements
Advertisements

Question

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

Advertisements

Solution

\[\lim_{x \to a} \left\{ \frac{\sin x}{\sin a} \right\}^\frac{1}{\left( x - a \right)} \]
\[ = \lim_{x \to a} \left\{ 1 + \frac{\sin x}{\sin a} - 1 \right\}^\frac{1}{\left( x - a \right)} \]
\[ = \lim_{x \to a} \left\{ 1 + \frac{\sin x - \sin a}{\sin a} \right\}^\frac{1}{\left( x - a \right)} \]
\[ = e {}^\lim_{x \to a} \left( \frac{\sin x - \sin a}{\sin a} \right) \times \frac{1}{\left( x - a \right)} \]
\[ = e {}^\lim_{x \to a} \left( \frac{2 \cos \left( \frac{x + a}{2} \right) \sin \left( \frac{x - a}{2} \right)}{\sin a \times \left( \frac{x - a}{2} \right) \times 2} \right) \]
\[ = e^\frac{\cos a}{\sin a} \]
` = e^\cot a`

shaalaa.com
  Is there an error in this question or solution?
Chapter 29: Limits - Exercise 29.12 [Page 77]

APPEARS IN

RD Sharma Mathematics [English] Class 11
Chapter 29 Limits
Exercise 29.12 | Q 9 | Page 77

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):}`


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


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


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


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


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


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


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


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


\[\lim_{x \to 0} \frac{\sqrt{1 + x^2} - \sqrt{1 + x}}{\sqrt{1 + x^3} - \sqrt{1 + x}}\] 


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


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


\[\lim_{x \to 0} \frac{\log \left( 1 + x \right)}{3^x - 1}\]


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


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


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


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


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


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


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


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


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


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


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


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


\[\lim_{x \to 0} \frac{3^{2 + x} - 9}{x}\]


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


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


Write the value of \[\lim_{x \to - \infty} \left( 3x + \sqrt{9 x^2 - x} \right) .\]


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


Evaluate: `lim_(h -> 0) (sqrt(x + h) - sqrt(x))/h`


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


Let f(x) be a polynomial of degree 4 having extreme values at x = 1 and x = 2. If `lim_(x rightarrow 0) ((f(x))/x^2 + 1)` = 3 then f(–1) is equal to ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×