हिंदी

Lim X → 0 3 2 + X − 9 X

Advertisements
Advertisements

प्रश्न

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

Advertisements

उत्तर

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

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 29: Limits - Exercise 29.1 [पृष्ठ ७२]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 29 Limits
Exercise 29.1 | Q 40 | पृष्ठ ७२

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):}`


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)`


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


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


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


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


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


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


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


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


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


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


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


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


\[\lim_{x \to 0} \frac{5^x + 3^x + 2^x - 3}{x}\]


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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×