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):}`


if `f(x) = { (mx^2 + n, x < 0),(nx + m, 0<= x <= 1),(nx^3 + m, x > 1):}`

For what integers m and n does `lim_(x-> 0) f(x)` and `lim_(x -> 1) f(x)` exist?


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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×