Advertisements
Advertisements
प्रश्न
\[\lim_{x \to 0} \frac{ax + b}{cx + d}, d \neq 0\]
Advertisements
उत्तर
\[ = \frac{a \times 0 + b}{c \times 0 + d}\]
\[ = \frac{b}{d}\]
APPEARS IN
संबंधित प्रश्न
\[\lim_{x \to 2} \left( 3 - x \right)\]
\[\lim_{x \to 3} \frac{x^2 - 9}{x + 2}\]
\[\lim_{x \to 1} \left( \frac{1}{x^2 + x - 2} - \frac{x}{x^3 - 1} \right)\]
\[\lim_{x \to 2} \left( \frac{1}{x - 2} - \frac{2}{x^2 - 2x} \right)\]
\[\lim_{x \to 1} \frac{x^4 - 3 x^3 + 2}{x^3 - 5 x^2 + 3x + 1}\]
\[\lim_{x \to 2} \frac{x^3 + 3 x^2 - 9x - 2}{x^3 - x - 6}\]
\[\lim_{x \to a} \frac{\left( x + 2 \right)^{5/2} - \left( a + 2 \right)^{5/2}}{x - a}\]
\[\lim_{x \to a} \frac{x^{5/7} - a^{5/7}}{x^{2/7} - a^{2/7}}\]
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.
\[\lim_{x \to \infty} \sqrt{x + 1} - \sqrt{x}\]
\[\lim_{n \to \infty} \left[ \frac{1^3 + 2^3 + . . . n^3}{\left( n - 1 \right)^4} \right]\]
\[\lim_{n \to \infty} \left[ \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + . . . + \frac{1}{3^n} \right]\]
\[\lim_{x \to \infty} \left[ \frac{x^4 + 7 x^3 + 46x + a}{x^4 + 6} \right]\] where a is a non-zero real number.
\[\lim_{x \to 0} \frac{\sin 5x}{\tan 3x}\]
\[\lim_{x \to 0} \frac{\sin^2 4 x^2}{x^4}\]
\[\lim_{x \to 0} \frac{x^2 + 1 - \cos x}{x \sin x}\]
\[\lim_{x \to 0} \frac{\sin \left( 3 + x \right) - \sin \left( 3 - x \right)}{x}\]
\[\lim_{x \to 0} \frac{1 - \cos 5x}{1 - \cos 6x}\]
\[\lim_{x \to \frac{\pi}{2}} \frac{\cos^2 x}{1 - \sin x}\]
\[\lim_{x \to a} \frac{\sin \sqrt{x} - \sin \sqrt{a}}{x - a}\]
\[\lim_{n \to \infty} 2^{n - 1} \sin \left( \frac{a}{2^n} \right)\]
Evaluate the following limit:
\[\lim_{x \to \pi} \frac{1 - \sin\frac{x}{2}}{\cos\frac{x}{2}\left( \cos\frac{x}{4} - \sin\frac{x}{4} \right)}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{{cosec}^2 x - 2}{\cot x - 1}\]
\[\lim_{x \to \frac{3\pi}{2}} \frac{1 + {cosec}^3 x}{\cot^2 x}\]
Write the value of \[\lim_{x \to \infty} \frac{\sin x}{x} .\]
Write the value of \[\lim_{x \to \pi} \frac{\sin x}{x - \pi} .\]
\[\lim_{x \to \infty} \frac{\sin x}{x}\] equals
\[\lim_{h \to 0} \left\{ \frac{1}{h\sqrt[3]{8 + h}} - \frac{1}{2h} \right\} =\]
\[\lim_{x \to 1} \frac{\sin \pi x}{x - 1}\]
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
\[\lim_{x \to \pi/4} \frac{4\sqrt{2} - \left( \cos x + \sin x \right)^5}{1 - \sin 2x}\] 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 \[\lim_{x \to \infty} \frac{n!}{\left( n + 1 \right)! - n!}\]
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 limits: `lim_(x -> 0)[(root(3)(1 + x) - sqrt(1 + x))/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 limits: `lim_(x -> 3) [sqrt(x + 6)/x]`
Evaluate the following limit.
`lim_(x->3)[sqrt(x + 6)/x]`
