Advertisements
Advertisements
Question
\[\lim_{x \to 0} \frac{\sec 5x - \sec 3x}{\sec 3x - \sec x}\]
Advertisements
Solution
\[\lim_{x \to 0} \left[ \frac{\sec 5x - \sec 3x}{\sec 3x - \sec x} \right]\]
\[= \lim_{x \to 0} \left[ \frac{\frac{1}{\cos 5x}\frac{1}{\cos 3x}}{\frac{1}{\cos 3x} - \frac{1}{\cos x}} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\cos 3x - \cos 5x}{\cos 5x \cos 3x\left\{ \frac{\cos x - \cos 3x}{\cos x \cos 3x} \right\}} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\left( \cos3x - \cos5x \right)\cos x}{\cos 5x\left\{ \cos x - \cos3x \right\}} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{- 2\sin\left( \frac{3x + 5x}{2} \right)\sin\left( \frac{3x - 5x}{2} \right) \times \cos x}{\cos\left( 5x \right)\left[ - 2\sin\left( \frac{x + 3x}{2} \right)\sin\left( \frac{x - 3x}{2} \right) \right]} \right] \left[ \because cosC - cosD = - 2\sin\left( \frac{C + D}{2} \right)\sin\left( \frac{C - D}{2} \right) \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\sin\left( 4x \right) \times \sin\left( - x \right) \times \cos x}{\cos\left( 5x \right) \times \sin\left( 2x \right) \times \sin\left( - x \right)} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\sin\left( 4x \right)}{4x} \times \frac{4x}{\frac{\sin\left( 2x \right)}{2x} \times 2x} \times \frac{\cos x}{\cos5x} \right]\]
\[ = \frac{4}{2}\frac{\cos0}{\cos0}\]
\[ = 2\]
\[\]
APPEARS IN
RELATED QUESTIONS
Suppose f(x) = `{(a+bx, x < 1),(4, x = 1),(b-ax, x > 1):}` and if `lim_(x -> 1) f(x) = f(1)` what are possible values of a and b?
\[\lim_{x \to 1} \frac{x^2 + 1}{x + 1}\]
\[\lim_{x \to 3} \frac{\sqrt{2x + 3}}{x + 3}\]
\[\lim_{x \to a} \frac{\sqrt{x} + \sqrt{a}}{x + a}\]
\[\lim_{x \to - 5} \frac{2 x^2 + 9x - 5}{x + 5}\]
\[\lim_{x \to 2} \left( \frac{1}{x - 2} - \frac{4}{x^3 - 2 x^2} \right)\]
\[\lim_{x \to 1} \frac{x^3 + 3 x^2 - 6x + 2}{x^3 + 3 x^2 - 3x - 1}\]
\[\lim_{x \to 1} \frac{\sqrt{x^2 - 1} + \sqrt{x - 1}}{\sqrt{x^2 - 1}}, x > 1\]
\[\lim_{x \to a} \frac{\left( x + 2 \right)^{3/2} - \left( a + 2 \right)^{3/2}}{x - a}\]
If \[\lim_{x \to a} \frac{x^3 - a^3}{x - a} = \lim_{x \to 1} \frac{x^4 - 1}{x - 1},\] find all possible values of a.
`lim_(x->∞) [x{sqrt(x^2+1) - sqrt(x^2-1)}]`
\[\lim_{n \to \infty} \left[ \frac{1^2 + 2^2 + . . . + n^2}{n^3} \right]\]
\[\lim_{x \to 0} \frac{2x - \sin x}{\tan x + x}\]
\[\lim_{x \to 0} \frac{\sin 3x - \sin x}{\sin x}\]
\[\lim_{x \to 0} \frac{\sin 2x \left( \cos 3x - \cos x \right)}{x^3}\]
\[\lim_{x \to 0} \frac{x^3 \cot x}{1 - \cos x}\]
\[\lim_{x \to 0} \frac{\sin \left( 3 + x \right) - \sin \left( 3 - x \right)}{x}\]
\[\lim_{x \to 0} \frac{\tan 2x - \sin 2x}{x^3}\]
\[\lim_{x \to 0} \left( cosec x - \cot x \right)\]
\[\lim_{x \to \pi} \frac{\sin x}{\pi - x}\]
\[\lim_{x \to \frac{\pi}{2}} \frac{\sin 2x}{\cos x}\]
\[\lim_{x \to a} \frac{\sin \sqrt{x} - \sin \sqrt{a}}{x - a}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2} - \cos x - \sin x}{\left( 4x - \pi \right)^2}\]
Write the value of \[\lim_{x \to 0^-} \left[ x \right] .\]
Write the value of \[\lim_{x \to 1^-} x - \left[ x \right] .\]
\[\lim_{x \to 0^-} \frac{\sin \left[ x \right]}{\left[ x \right]} .\]
\[\lim_{x \to 0} \frac{\sqrt{1 - \cos 2x}}{x} .\]
Write the value of \[\lim_{x \to \pi} \frac{\sin x}{x - \pi} .\]
Write the value of \[\lim_{x \to 0} \frac{\sin x^\circ}{x} .\]
\[\lim_{x \to a} \frac{x^n - a^n}{x - a}\] is equal at
\[\lim_{x \to \pi/4} \frac{\sqrt{2} \cos x - 1}{\cot x - 1}\] is equal to
\[\lim_{n \to \infty} \left\{ \frac{1}{1 . 3} + \frac{1}{3 . 5} + \frac{1}{5 . 7} + . . . + \frac{1}{\left( 2n + 1 \right) \left( 2n + 3 \right)} \right\}\]is equal to
If \[\lim_{x \to 1} \frac{x + x^2 + x^3 + . . . + x^n - n}{x - 1} = 5050\] then n equal
The value of \[\lim_{x \to \infty} \frac{\sqrt{1 + x^4} + \left( 1 + x^2 \right)}{x^2}\] is
\[\lim_{x \to \pi/3} \frac{\sin \left( \frac{\pi}{3} - x \right)}{2 \cos x - 1}\] is equal to
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]`
Evaluate the following limit.
`lim_(x->3)[sqrt(x + 6)/x]`
Evaluate the following limit.
`lim_(x->5)[(x^3 -125)/(x^5 - 3125)]`
Evaluate the Following limit:
`lim_(x->7)[((root(3)(x)-root(3)(7))(root(3)(x)+root(3)(7)))/(x-7)]`
