हिंदी

Lim X → a { Sin X Sin a } 1 X − a

Advertisements
Advertisements

प्रश्न

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

Advertisements

उत्तर

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

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 29: Limits - Exercise 29.12 [पृष्ठ ७७]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 29 Limits
Exercise 29.12 | Q 9 | पृष्ठ ७७

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

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):}`


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)`


If f(x) = `{(|x| +  1,x < 0), (0, x = 0),(|x| -1, x > 0):}`

For what value (s) of a does `lim_(x -> a)`  f(x) exists?


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


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


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


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


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


\[\lim_{x \to 5} \frac{x - 5}{\sqrt{6x - 5} - \sqrt{4x + 5}}\] 


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


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


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


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


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


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


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


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


\[\lim_{x \to 0} \frac{5^x - 1}{\sqrt{4 + x} - 2}\]


\[\lim_{x \to 0} \frac{a^x + b^x + c^x - 3}{x}\] 


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


\[\lim_{x \to \infty} \left( a^{1/x} - 1 \right)x\]


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


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


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


\[\lim_{x \to 0} \frac{\log \left| 1 + x^3 \right|}{\sin^3 x}\] 

 


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


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


\[\lim_{x \to 0} \left\{ \frac{e^x + e^{- x} - 2}{x^2} \right\}^{1/ x^2}\]


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


\[\lim_{x \to 0} \left\{ \frac{e^x + e^{- x} - 2}{x^2} \right\}^{1/ x^2}\]


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


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×