हिंदी

Lim X → ∞ { X 2 + 2 X + 3 2 X 2 + X + 5 } 3 X − 2 3 X + 2

Advertisements
Advertisements

प्रश्न

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

Advertisements

उत्तर

\[\lim_{x \to \infty} \left( \frac{x^2 + 2x + 3}{2 x^2 + x + 5} \right)^\left( \frac{3x - 2}{3x + 2} \right) \]
\[ = \lim_{x \to \infty} \left[ 1 + \frac{x^2 + 2x + 3}{2 x^2 + x + 5} - 1 \right]\left( {}^\frac{3x - 2}{3x + 2} \right)\]
\[ = \lim_{x \to \infty} \left[ 1 + \frac{\left( x^2 + 2x + 3 \right) - \left( 2 x^2 + x + 5 \right)}{2 x^2 + x + 5} \right]^\left( \frac{3x - 2}{3x + 2} \right) \]
\[ = \lim_{x \to \infty} \left[ 1 + \frac{\left( - x^2 + x - 2 \right)}{2 x^2 + x + 5} \right]^\left( \frac{3x - 2}{3x + 2} \right) \]
\[ = e^\lim_{x \to \infty} \left( \frac{- x^2 + x - 2}{2 x^2 + x + 5} \right) \times \left( \frac{3x - 2}{3x + 2} \right) \]
\[ = e^\lim_{x \to \infty} \left( \frac{- 1 + \frac{1}{x} - \frac{2}{x^2}}{2 + \frac{1}{x} + \frac{5}{x^2}} \right) \times \left( \frac{3 - \frac{2}{x}}{3 + \frac{2}{x}} \right) \]
\[ = e^{- \frac{1}{2} \times 1} \]
\[ = \frac{1}{\sqrt{e}}\]
\[\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 29: Limits - Exercise 29.11 [पृष्ठ ७७]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 29 Limits
Exercise 29.11 | Q 6 | पृष्ठ ७७

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

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


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


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


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


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


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


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


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


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


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


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


\[\lim_{x \to \sqrt{6}} \frac{\sqrt{5 + 2x} - \left( \sqrt{3} + \sqrt{2} \right)}{x^2 - 6}\] 

 


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×