मराठी

Lim X → 0 √ a 2 + X 2 − a X 2 - Mathematics

Advertisements
Advertisements

प्रश्न

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

Advertisements

उत्तर

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


On putting x = 0 in the expression  \[\sqrt{a^2 + x^2} - a\] it becomes \[\frac{0}{0} .\] Rationalising the numerator: 

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

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

=  \[\frac{1}{\sqrt{a^2} + a}\] 

=  \[\frac{1}{2a}\] 

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 29: Limits - Exercise 29.4 [पृष्ठ २८]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
पाठ 29 Limits
Exercise 29.4 | Q 3 | पृष्ठ २८

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):}`


Find `lim_(x -> 0)` f(x), where `f(x) = {(x/|x|, x != 0),(0, x = 0):}`


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


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?


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


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


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


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


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


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


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


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


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


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


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


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


\[\lim_{x \to 0} \frac{8^x - 4^x - 2^x + 1}{x^2}\]


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


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


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


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


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


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


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


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


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


\[\lim_{x \to 5} \frac{e^x - e^5}{x - 5}\]


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


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


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


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


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 \[\lim_{x \to \pi/2} \frac{2x - \pi}{\cos x} .\] 


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


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×