Advertisements
Advertisements
प्रश्न
\[\lim_{x \to 0} \frac{\cos ax - \cos bx}{\cos cx - \cos dx}\]
Advertisements
उत्तर
\[\lim_{x \to 0} \left[ \frac{\cos ax - \cos bx}{\cos cx - \cos dx} \right]\] It is of the form \[\left( \frac{0}{0} \right)\]
\[\Rightarrow \lim_{x \to 0} \left[ \frac{- 2\sin\left( \frac{ax + bx}{2} \right)\sin\left( \frac{ax - bx}{2} \right)}{- 2\sin\left( \frac{cx + dx}{a2} \right)\sin\left( \frac{cx - dx}{2} \right)} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\frac{\sin\left( \frac{ax + bx}{2} \right)}{\left( \frac{ax + b}{x} \right)} \times \left( \frac{ax + bx}{2} \right) \times \frac{\sin\left( \frac{ax - bx}{2} \right)}{\left( \frac{ax - bx}{2} \right)} \times \left( \frac{ax - bx}{2} \right)}{\frac{\sin\left( \frac{cx + dx}{2} \right)}{\left( \frac{cx + dx}{2} \right)} \times \left( \frac{cx + dx}{2} \right) \times \frac{\sin\left( \frac{cx - dx}{2} \right)}{\left( \frac{cx - dx}{2} \right)} \times \left( \frac{cx - dx}{2} \right)} \right]\]
\[ = 1 \times \lim_{x \to 0} \left[ \frac{\left( \frac{ax + bx}{2} \right)\left( \frac{ax - bx}{2} \right)}{\left( \frac{cx + dx}{2} \right)\left( \frac{cx - dx}{2} \right)} \right]\]
\[ = \lim_{x \to 0} \frac{x^2 \left( a + b \right)\left( a - b \right)}{x^2 \left( c + d \right)\left( c - d \right)}\]
\[ = \frac{a^2 - b^2}{c^2 - d^2}\]
APPEARS IN
संबंधित प्रश्न
\[\lim_{x \to 0} \frac{3x + 1}{x + 3}\]
\[\lim_{x \to - 5} \frac{2 x^2 + 9x - 5}{x + 5}\]
\[\lim_{x \to \sqrt{3}} \frac{x^4 - 9}{x^2 + 4\sqrt{3}x - 15}\]
\[\lim_{x \to 2} \left( \frac{1}{x - 2} - \frac{2}{x^2 - 2x} \right)\]
\[\lim_{x \to 2} \frac{x^3 + 3 x^2 - 9x - 2}{x^3 - x - 6}\]
\[\lim_{x \to 0} \frac{\left( 1 + x \right)^6 - 1}{\left( 1 + x \right)^2 - 1}\]
\[\lim_{x \to - 1/2} \frac{8 x^3 + 1}{2x + 1}\]
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.
\[\lim_{x \to \infty} \sqrt{x^2 + 7x - x}\]
\[\lim_{x \to \infty} \frac{3 x^{- 1} + 4 x^{- 2}}{5 x^{- 1} + 6 x^{- 2}}\]
\[\lim_{x \to \infty} \left[ \sqrt{x}\left\{ \sqrt{x + 1} - \sqrt{x} \right\} \right]\]
\[\lim_{x \to 0} \frac{x^2}{\sin x^2}\]
\[\lim_{x \to 0} \left[ \frac{x^2}{\sin x^2} \right]\]
\[\lim_{x \to 0} \frac{\tan^2 3x}{x^2}\]
\[\lim_{x \to 0} \frac{\sin \left( 2 + x \right) - \sin \left( 2 - x \right)}{x}\]
\[\lim_{x \to 0} \frac{\sin 2x \left( \cos 3x - \cos x \right)}{x^3}\]
\[\lim_{x \to 0} \frac{x^3 \cot x}{1 - \cos x}\]
\[\lim_{x \to 0} \frac{\sin ax + bx}{ax + \sin bx}\]
Evaluate the following limit:
\[\lim_{h \to 0} \frac{\left( a + h \right)^2 \sin\left( a + h \right) - a^2 \sin a}{h}\]
Evaluate the following limit:
\[\lim_{x \to \frac{\pi}{3}} \frac{\sqrt{1 - \cos6x}}{\sqrt{2}\left( \frac{\pi}{3} - x \right)}\]
\[\lim_{x \to \frac{\pi}{8}} \frac{\cot 4x - \cos 4x}{\left( \pi - 8x \right)^3}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2} - \cos x - \sin x}{\left( 4x - \pi \right)^2}\]
\[\lim_{x \to \pi} \frac{1 + \cos x}{\tan^2 x}\]
\[\lim_{x \to 0} \frac{8^x - 2^x}{x}\]
\[\lim_{x \to 0} \left( \cos x \right)^{1/\sin x}\]
\[\lim_{n \to \infty} \left\{ \frac{1}{1 - n^2} + \frac{2}{1 - n^2} + . . . + \frac{n}{1 - n^2} \right\}\]
If \[\lim_{x \to 1} \frac{x + x^2 + x^3 + . . . + x^n - n}{x - 1} = 5050\] then n equal
If \[f\left( x \right) = \left\{ \begin{array}{l}x \sin \frac{1}{x}, & x \neq 0 \\ 0, & x = 0\end{array}, \right.\] then \[\lim_{x \to 0} f\left( x \right)\] equals
\[\lim_{x \to 2} \frac{\sqrt{1 + \sqrt{2 + x} - \sqrt{3}}}{x - 2}\] is equal to
\[\lim_{x \to 0} \frac{8}{x^8}\left\{ 1 - \cos \frac{x^2}{2} - \cos \frac{x^2}{4} + \cos \frac{x^2}{2} \cos \frac{x^2}{4} \right\}\] is equal to
The value of \[\lim_{x \to \pi/2} \left( \sec x - \tan x \right)\]is
The value of \[\lim_{x \to \infty} \frac{n!}{\left( n + 1 \right)! - n!}\]
\[\lim_{x \to 0} \frac{\left| \sin x \right|}{x}\]
Evaluate the following limit:
`lim_(x -> 5)[(x^3 - 125)/(x^5 - 3125)]`
Evaluate `lim_(h -> 0) ((a + h)^2 sin (a + h) - a^2 sina)/h`
Number of values of x where the function
f(x) = `{{:((tanxlog(x - 2))/(x^2 - 4x + 3); x∈(2, 4) - {3, π}),(1/6tanx; x = 3"," π):}`
is discontinuous, is ______.
Evaluate the following limit:
`lim_(x->3)[(sqrt(x+6))/x]`
Evaluate the following limit:
`lim_(x->7)[((root(3)(x)-root(3)(7))(root(3)(x)+root(3)(7)))/(x-7)]`
Evaluate the following limit:
`lim_(x->7)[((root(3)(x) - root(3)(7))(root(3)(x) + root(3)(7)))/(x - 7)]`
