मराठी

Lim X → 0 9 X − 2 . 6 X + 4 X X 2 - Mathematics

Advertisements
Advertisements

प्रश्न

\[\lim_{x \to 0} \frac{9^x - 2 . 6^x + 4^x}{x^2}\] 

उत्तर

\[\lim_{x \to 0} \left[ \frac{9^x - 2 . 6^x + 4^x}{x^2} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\left( 3^x \right)^2 - 2 \cdot 3^x \cdot 2^x + \left( 2^x \right)^2}{x^2} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\left( 3^x - 2^x \right)^2}{x^2} \right]\]
\[ = \lim_{x \to 0} \left[ \left( \frac{3^x - 2^x}{2^x} \right)^2 \times \frac{\left( 2^x \right)^2}{x^2} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\left( \frac{3}{2} \right)^x - 1}{x} \right]^2 \times 2^{2x} \]
\[ = \left[ \log \left( \frac{3}{2} \right) \right]^2 \times 2^0 \]
\[ = \left[ \log \left( \frac{3}{2} \right) \right]^2\]

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 6 | पृष्ठ ७१

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

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?


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


\[\lim_{x \to \infty} \left\{ \frac{3 x^2 + 1}{4 x^2 - 1} \right\}^\frac{x^3}{1 + x}\]


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


Write the value of \[\lim_{n \to \infty} \frac{1 + 2 + 3 + . . . + n}{n^2} .\]


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×