मराठी

Lim X → 0 5 X − 1 √ 4 + X − 2 - Mathematics

Advertisements
Advertisements

प्रश्न

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

Advertisements

उत्तर

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

Rationalising the denominator, we get: 

\[= \lim_{x \to 0} \left[ \frac{\left( 5^x - 1 \right) \left( \sqrt{4 + x} + 2 \right)}{\left( \sqrt{4 + x} - 2 \right) \left( \sqrt{4 + x} + 2 \right)} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\left( 5^x - 1 \right) \left( \sqrt{4 + x} + 2 \right)}{4 + x - 4} \right]\]
\[ = \lim_{x \to 0} \left[ \left( \frac{5^x - 1}{x} \right) \left( \sqrt{4 + x} + 2 \right) \right] \left\{ \because \lim_{x \to 0} \left( \frac{a^x - 1}{x} \right) = \log a \right\}\]
\[ = \log 5 \times \left( \sqrt{4 + 0} + 2 \right)\]
\[ = 4 \log 5\]

 

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 29: Limits - Exercise 29.1 [पृष्ठ ७१]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
पाठ 29 Limits
Exercise 29.1 | Q 1 | पृष्ठ ७१

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?


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


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


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


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


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


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


\[\lim_{x \to 7} \frac{4 - \sqrt{9 + x}}{1 - \sqrt{8 - x}}\] 


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


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


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


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


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


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


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


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


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


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


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


\[\lim_{x \to 0} \frac{\log \left| 1 + x^3 \right|}{\sin^3 x}\] 

 


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


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


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


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


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


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


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


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×