Advertisements
Advertisements
प्रश्न
\[\lim_{x \to 0} \left( \cos x + a \sin bx \right)^{1/x}\]
Advertisements
उत्तर
\[\lim_{x \to 0} \left( \cos x + a \sin bx \right)^\frac{1}{x} \]
\[ = \lim_{x \to 0} \left[ 1 + \cos x + a \sin bx - 1 \right]^\frac{1}{x} \]
\[\text{ Using the theoremgiven below }:\]
\[If \lim_{x \to a} f\left( x \right) = \lim_{x \to a} g\left( x \right) = 0 \text{ such that } \lim_{x \to a} \frac{f\left( x \right)}{g\left( x \right)} \text{ exists, then } \lim_{x \to a} \left[ 1 + f\left( x \right) \right]^\frac{1}{g\left( x \right)} = e^\lim_{x \to a} \frac{f\left( x \right)}{g\left( x \right)} . \]
\[Here: \]
\[ f\left( x \right) = \cos x + a \sin bx - 1\]
\[ g\left( x \right) = x\]
\[ \Rightarrow e^\lim_{x \to 0} \left[ \frac{\cos x + a \sin bx - 1}{x} \right] \]
\[ = e^\lim_{x \to 0} \left[ \frac{b \times a \sin bx}{bx} - \frac{\left( 1 - \cos x \right)}{x} \right] \]
\[ = e^\lim_{x \to 0} \left( \frac{ab \sin bx}{bx} - \frac{2 \sin^2 \frac{x}{2}}{x} \right) \]
\[ = e^{ab}\]
APPEARS IN
संबंधित प्रश्न
\[\lim_{x \to a} \frac{\sqrt{x} + \sqrt{a}}{x + a}\]
\[\lim_{x \to 2} \left( 3 - x \right)\]
\[\lim_{x \to - 1/2} \frac{8 x^3 + 1}{2x + 1}\]
\[\lim_{x \to - 1} \frac{x^3 + 1}{x + 1}\]
\[\lim_{x \to 0} \frac{\left( 1 + x \right)^6 - 1}{\left( 1 + x \right)^2 - 1}\]
\[\lim_{x \to a} \frac{x^{2/7} - a^{2/7}}{x - a}\]
`lim_(x->∞) [x{sqrt(x^2+1) - sqrt(x^2-1)}]`
Evaluate: \[\lim_{n \to \infty} \frac{1^4 + 2^4 + 3^4 + . . . + n^4}{n^5} - \lim_{n \to \infty} \frac{1^3 + 2^3 + . . . + n^3}{n^5}\]
\[\lim_{x \to 0} \frac{\sin x \cos x}{3x}\]
\[\lim_{x \to 0} \frac{\sin x^n}{x^n}\]
\[\lim_{x \to 0} \frac{1 - \cos mx}{x^2}\]
\[\lim_{x \to 0} \frac{\sin x^2 \left( 1 - \cos x^2 \right)}{x^6}\]
\[\lim_{x \to 0} \frac{\sin^2 4 x^2}{x^4}\]
\[\lim_{x \to 0} \frac{\sin \left( 2 + x \right) - \sin \left( 2 - x \right)}{x}\]
\[\lim_{x \to 0} \frac{x \tan x}{1 - \cos x}\]
\[\lim_{x \to 0} \frac{ax + x \cos x}{b \sin x}\]
\[\lim_{x \to 0} \frac{cosec x - \cot x}{x}\]
\[\lim_{x \to \frac{\pi}{2}} \frac{\sin 2x}{\cos x}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{\cos x} - \sqrt{\sin x}}{x - \frac{\pi}{4}}\]
\[\lim_{x \to \pi} \frac{\sqrt{5 + \cos x} - 2}{\left( \pi - x \right)^2}\]
\[\lim_{x \to a} \frac{\cos \sqrt{x} - \cos \sqrt{a}}{x - a}\]
\[\lim_{x \to 1} \frac{1 - x^2}{\sin \pi x}\]
\[\lim_{x \to 2} \frac{x^2 - x - 2}{x^2 - 2x + \sin \left( x - 2 \right)}\]
\[\lim_{x \to 0} \frac{\log \left( a + x \right) - \log a}{x}\]
Write the value of \[\lim_{x \to 0^+} \left[ x \right] .\]
\[\lim_{x \to \pi} \frac{\sin x}{x - \pi} .\]
Write the value of \[\lim_{x \to 0} \frac{\sin x^\circ}{x} .\]
If \[f\left( x \right) = x \sin \left( 1/x \right), x \neq 0,\] then \[\lim_{x \to 0} f\left( x \right) =\]
\[\lim_{x \to 0} \frac{\sin x^0}{x}\]
The value of \[\lim_{x \to \pi/2} \left( \sec x - \tan x \right)\]is
If \[f\left( x \right) = \begin{cases}\frac{\sin\left[ x \right]}{\left[ x \right]}, & \left[ x \right] \neq 0 \\ 0, & \left[ x \right] = 0\end{cases}\] where denotes the greatest integer function, then \[\lim_{x \to 0} f\left( x \right)\]
Evaluate the following limits: `lim_(x -> 0)[(root(3)(1 + x) - sqrt(1 + x))/x]`
Evaluate the following limits: `lim_(x -> 5)[(x^3 - 125)/(x^2 - 25)]`
Which of the following function is not continuous at x = 0?
Evaluate: `lim_(x -> 1) ((1 + x)^6 - 1)/((1 + x)^2 - 1)`
Evaluate the following limit :
`lim_(x->7)[[(root3(x)- root3(7))(root3(x) + root3(7)))/(x-7)]`
Evaluate the following limit:
`lim_(x->3)[(sqrt(x+6))/x]`
Evaluate the following limit:
`lim_(x->3)[sqrt(x+6)/x]`
Evaluate the following limit:
`\underset{x->3}{lim}[sqrt(x +6)/(x)]`
