हिंदी

Lim X → 4 2 − √ X 4 − X - Mathematics

Advertisements
Advertisements

प्रश्न

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

उत्तर

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

=  \[\lim_{x \to 4} \left[ \frac{2 - \sqrt{x}}{2 - \left( \sqrt{x} \right)^2} \right]\]

= \[\lim_{x \to 4} \left[ \frac{\left( 2 - \sqrt{x} \right)}{\left( 2 - \sqrt{x} \right)\left( 2 + \sqrt{x} \right)} \right]\] 

= \[\frac{1}{2 + \sqrt{4}}\]

= \[\frac{1}{2 + 2}\] 

=  \[\frac{1}{4}\]

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 23 | पृष्ठ २९

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

If f(x) = `{(|x| +  1,x < 0), (0, x = 0),(|x| -1, x > 0):}`

For what value (s) of a does `lim_(x -> a)`  f(x) exists?


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?


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×