मराठी

Lim X → √ 10 √ 7 + 2 X − ( √ 5 + √ 2 ) X 2 − 10

Advertisements
Advertisements

प्रश्न

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

Advertisements

उत्तर

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

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

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

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

Rationalising the numerator:

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

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

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

=  \[\frac{2}{\left( \sqrt{10} + \sqrt{10} \right) \left( 2\sqrt{7 + 2\sqrt{10}} \right)}\] 

\[= \frac{2}{\left( 2\sqrt{10} \right) \times 2\sqrt{7 + 2\sqrt{10}}}\]
\[ = \frac{1}{2\sqrt{10}\sqrt{\left( \sqrt{5} + \sqrt{2} \right)^2}}\] 

=  \[\frac{1}{2\sqrt{10}\left( \sqrt{5} + \sqrt{2} \right)} \times \frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} - \sqrt{2}}\] 

\[= \frac{1}{2\sqrt{10}}\left[ \frac{\sqrt{5} - \sqrt{2}}{\left( \sqrt{5} \right)^2 - \left( \sqrt{2} \right)^2} \right]\]
= \[\frac{\sqrt{5} - \sqrt{2}}{6\sqrt{10}}\] 

 

 

 

 
shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 29: Limits - Exercise 29.4 [पृष्ठ २९]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 11
पाठ 29 Limits
Exercise 29.4 | Q 32 | पृष्ठ २९

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):}`


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 the function f(x) satisfies `lim_(x -> 1) (f(x) - 2)/(x^2 - 1) = pi`, evaluate `lim_(x -> 1) f(x)`.


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×