हिंदी

lim x → 0 √ 2 − x − √ 2 + x x - Mathematics

Advertisements
Advertisements

प्रश्न

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

उत्तर

\[\lim_{x \to 0} \left[ \frac{\sqrt{2 - x} - \sqrt{2 + x}}{x} \right]\] It is of the form  \[\frac{0}{0}\]

Rationalising the numerator: 

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

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

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

= \[\frac{- 2}{2\sqrt{2}}\] 

=  \[- \frac{1}{\sqrt{2}}\] 

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 29: Limits - Exercise 29.4 [पृष्ठ २९]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
अध्याय 29 Limits
Exercise 29.4 | Q 26 | पृष्ठ २९

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):}`


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)`


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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

 


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


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


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


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


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


\[\lim_{x \to 0} \frac{e^{bx} - e^{ax}}{x} \text{ where } 0 < a < b\] 


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


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


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


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


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


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×