Advertisements
Advertisements
Question
Evaluate the following limit:
\[\lim_{x \to 0} \frac{\sin\left( \alpha + \beta \right)x + \sin\left( \alpha - \beta \right)x + \sin2\alpha x}{\cos^2 \beta x - \cos^2 \alpha x}\]
Advertisements
Solution
\[\lim_{x \to 0} \frac{\sin\left( \alpha + \beta \right)x + \sin\left( \alpha - \beta \right)x + \sin2\alpha x}{\cos^2 \beta x - \cos^2 \alpha x}\]
`therefore cos^2 beta x - cos^2alpha x`
`= cos^2betax - cos^2alphax + cos^2betax * cos^2alphax - cos^2betax * cos^2alphax`
= `cos^2 beta x - cos^2 beta x * cos^2 alpha x - cos^2 alpha x + cos^2 beta x * cos^2 alpha x`
= `cos^2 beta x (1 - cos^2 alpha x) - cos^2 alpha x (1 - cos^2 beta x)`
= `cos^2 beta x * sin^2 alpha x - cos^2 alpha x * sin^2 beta x`
= `(cos beta x * sin alpha x)^2 - (cos alpha x * sin beta x)`
= `(cos beta x * sin alpha x + cos alpha x * sin beta x)(cos beta x * sin alpha x - cos alpha x * sin beta x)`
= `sin (alpha x + beta x) sin (alphax - betax)` ...[sin (A ± B) = sin A cos B ± cos A sin B]
= `sin (alpha + beta)x sin (alpha - beta)x`
`therefore lim_(x->0) (x{sin (alpha + beta )x + sin (alpha - beta)x + sin 2 alpha x})/(sin (alpha + beta)x sin (alpha - beta) x)`
`lim_(x->0) ((x{sin (alpha + beta )x + sin (alpha - beta)x + sin 2 alpha x})/x^2)/((sin (alpha + beta)x sin (alpha - beta) x)/x^2)`
`lim_(x->0) ((sin (alpha + beta)x)/x + (sin (alpha - beta)x)/x + (sin 2 alpha x)/x)/((sin (alpha + beta)x)/x xx (sin (alpha - beta)x)/x)`
`lim_(x->0) ((sin (alpha + beta)x)/((alpha + beta) x) xx (alpha + beta) + (sin (alpha - beta)x)/((alpha - beta)x) xx (alpha - beta) + (sin 2 alpha x)/(2alphax) xx 2alpha)/((sin (alpha + beta)x)/((alpha + beta)x) xx (alpha + beta) xx (sin (alpha - beta)x)/((alpha - beta)x) xx (alpha - beta))`
`= (alpha + beta + alpha - beta + 2alpha)/((alpha + beta)(alpha - beta))`
`= (4alpha)/(alpha^2 - beta^2)`
APPEARS IN
RELATED QUESTIONS
\[\lim_{x \to 0} \frac{3x + 1}{x + 3}\]
\[\lim_{x \to 2} \frac{x^3 - 8}{x^2 - 4}\]
\[\lim_{x \to 1} \frac{1 - x^{- 1/3}}{1 - x^{- 2/3}}\]
\[\lim_{x \to 1} \left\{ \frac{x - 2}{x^2 - x} - \frac{1}{x^3 - 3 x^2 + 2x} \right\}\]
\[\lim_{x \to a} \frac{x^{5/7} - a^{5/7}}{x^{2/7} - a^{2/7}}\]
\[\lim_{x \to 4} \frac{x^3 - 64}{x^2 - 16}\]
\[\lim_{x \to \infty} \sqrt{x^2 + cx - x}\]
\[\lim_{x \to 0} \frac{\sin 3x}{5x}\]
\[\lim_{x \to 0} \frac{3 \sin x - 4 \sin^3 x}{x}\]
\[\lim_{x \to 0} \frac{\sin 5x}{\tan 3x}\]
\[\lim_{x \to 0} \frac{7x \cos x - 3 \sin x}{4x + \tan x}\]
\[\lim_{x \to 0} \frac{\sec 5x - \sec 3x}{\sec 3x - \sec x}\]
\[\lim_{x \to 0} \frac{x \tan x}{1 - \cos 2x}\]
\[\lim_{x \to 0} \frac{1 - \cos 2x}{3 \tan^2 x}\]
\[\lim_\theta \to 0 \frac{\sin 4\theta}{\tan 3\theta}\]
\[\lim_{x \to 0} \frac{cosec x - \cot x}{x}\]
\[\lim_{x \to 0} \frac{3 \sin x - \sin 3x}{x^3}\]
\[\lim_{x \to \frac{\pi}{8}} \frac{\cot 4x - \cos 4x}{\left( \pi - 8x \right)^3}\]
\[\lim_{x \to 1} \frac{1 - x^2}{\sin \pi x}\]
\[\lim_{x \to \pi} \frac{\sqrt{2 + \cos x} - 1}{\left( \pi - x \right)^2}\]
Evaluate the following limit:
\[\lim_{x \to \pi} \frac{1 - \sin\frac{x}{2}}{\cos\frac{x}{2}\left( \cos\frac{x}{4} - \sin\frac{x}{4} \right)}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{2 - {cosec}^2 x}{1 - \cot x}\]
\[\lim_{x \to 0} \frac{8^x - 2^x}{x}\]
Write the value of \[\lim_{x \to 0^-} \left[ x \right] .\]
Write the value of \[\lim_{x \to 0^-} \left[ x \right] .\]
\[\lim_{x \to a} \frac{x^n - a^n}{x - a}\] is equal at
\[\lim_{x \to 0} \frac{8}{x^8}\left\{ 1 - \cos \frac{x^2}{2} - \cos \frac{x^2}{4} + \cos \frac{x^2}{2} \cos \frac{x^2}{4} \right\}\] is equal to
The value of \[\lim_{x \to \infty} \frac{n!}{\left( n + 1 \right)! - n!}\]
The value of \[\lim_{n \to \infty} \frac{\left( n + 2 \right)! + \left( n + 1 \right)!}{\left( n + 2 \right)! - \left( n + 1 \right)!}\] is
Evaluate the following limit:
`lim_(x -> 7)[((root(3)(x) - root(3)(7))(root(3)(x) + root(3)(7)))/(x - 7)]`
Evaluate the following limits: `lim_(x -> 5)[(x^3 - 125)/(x^2 - 25)]`
if `lim_(x -> 2) (x^"n"- 2^"n")/(x - 2)` = 80 then find the value of n.
Evaluate the following Limit:
`lim_(x -> 0) ((1 + x)^"n" - 1)/x`
If the value of `lim_(x -> 1) (1 - (1 - x))^"m"/x` is 99, then n = ______.
If `f(x) = {{:(x + 2",", x ≤ - 1),(cx^2",", x > -1):}`, find 'c' if `lim_(x -> -1) f(x)` exists
If f(x) = `{{:(1 if x "is rational"),(-1 if x "is rational"):}` is continuous on ______.
Evaluate the following limit :
`lim_(x->3)[sqrt(x+6)/x]`
