हिंदी

Lim X → 0 a X + B X − 2 X

Advertisements
Advertisements

प्रश्न

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

Advertisements

उत्तर

\[\lim_{x \to 0} \left[ \frac{a^x + b^x - 2}{x} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{a^x - 1 + b^x - 1}{x} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{a^x - 1}{x} \right] + \lim_{x \to 0} \left[ \frac{b^x - 1}{x} \right]\]
\[ = \log a + \log b\]
\[ = \log \left( ab \right)\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 29: Limits - Exercise 29.1 [पृष्ठ ७१]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 29 Limits
Exercise 29.1 | Q 5 | पृष्ठ ७१

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

Find `lim_(x -> 1)` f(x), where `f(x) = {(x^2 -1, x <= 1), (-x^2 -1, x > 1):}`


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


Let a1, a2,..., an be fixed real numbers and define a function f ( x) = ( x − a1 ) ( x − a2 )...( x − an ).

What is `lim_(x -> a_1) f(x)` ? For some a ≠ a1, a2, ..., an, compute `lim_(x -> a) f(x)`


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


`\lim_{x \to 0} \frac{e^\tan x - 1}{x}`


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


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


Write the value of \[\lim_{x \to - \infty} \left( 3x + \sqrt{9 x^2 - x} \right) .\]


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


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×