Advertisements
Advertisements
प्रश्न
If `f(x) = {{:(x + 2",", x ≤ - 1),(cx^2",", x > -1):}`, find 'c' if `lim_(x -> -1) f(x)` exists
Advertisements
उत्तर
Given, `f(x) = {{:(x + 2",", x ≤ - 1),(cx^2",", x > -1):}`
L.H.L = `lim_(x -> -1^-) f(x)`
= `lim_(x -> -1^-) (x + 2)`
= `lim(h -> 0) (-1 - h + 2)`
= `lim_(h -> 0) (1 - h)` = 1
R.H.L = `lim_(x -> 1^+) cx^2`
= `lim_(h -> 0) c(-1 + h)^2` = c
Since the limits exist.
∴ L.H.L = R.H.L
∴ c = 1
APPEARS IN
संबंधित प्रश्न
\[\lim_{x \to a} \frac{\sqrt{x} + \sqrt{a}}{x + a}\]
\[\lim_{x \to 2} \left( 3 - x \right)\]
\[\lim_{x \to 3} \frac{x^2 - 9}{x + 2}\]
\[\lim_{x \to \sqrt{3}} \frac{x^2 - 3}{x^2 + 3 \sqrt{3}x - 12}\]
\[\lim_{x \to 1} \frac{1 - x^{- 1/3}}{1 - x^{- 2/3}}\]
\[\lim_{x \to 0} \frac{\left( 1 + x \right)^6 - 1}{\left( 1 + x \right)^2 - 1}\]
If \[\lim_{x \to 3} \frac{x^n - 3^n}{x - 3} = 108,\] find the value of n.
\[\lim_{x \to \infty} \frac{3 x^3 - 4 x^2 + 6x - 1}{2 x^3 + x^2 - 5x + 7}\]
\[\lim_{x \to \infty} \sqrt{x + 1} - \sqrt{x}\]
\[\lim_{x \to 0} \frac{\tan 8x}{\sin 2x}\]
\[\lim_{x \to 0} \frac{\cos ax - \cos bx}{\cos cx - \cos dx}\]
\[\lim_\theta \to 0 \frac{\sin 3\theta}{\tan 2\theta}\]
\[\lim_{x \to 0} \frac{1 - \cos 2x}{\cos 2x - \cos 8x}\]
\[\lim_{x \to 0} \frac{x^3 \cot x}{1 - \cos x}\]
\[\lim_{x \to 0} \frac{\sin \left( 3 + x \right) - \sin \left( 3 - x \right)}{x}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}{x}\]
\[\lim_{x \to 0} \frac{x \cos x + \sin x}{x^2 + \tan x}\]
\[\lim_{x \to 0} \frac{\sin 3x + 7x}{4x + \sin 2x}\]
\[\lim_{x \to 0} \frac{5x + 4 \sin 3x}{4 \sin 2x + 7x}\]
\[\lim_{x \to 0} \frac{\sin ax + bx}{ax + \sin bx}\]
Evaluate the following limits:
\[\lim_{x \to 0} \frac{\cos ax - \cos bx}{\cos cx - 1}\]
\[\lim_{x \to \frac{\pi}{2}} \frac{\sin 2x}{\cos x}\]
\[\lim_{x \to \frac{\pi}{2}} \frac{1 - \sin x}{\left( \frac{\pi}{2} - x \right)^2}\]
\[\lim_{x \to a} \frac{\cos \sqrt{x} - \cos \sqrt{a}}{x - a}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{1 - \tan x}{1 - \sqrt{2} \sin x}\]
\[\lim_{x \to 0} \frac{\log \left( a + x \right) - \log a}{x}\]
Write the value of \[\lim_{x \to 1^-} x - \left[ x \right] .\]
Write the value of \[\lim_{x \to 0^-} \left[ x \right] .\]
\[\lim_{h \to 0} \left\{ \frac{1}{h\sqrt[3]{8 + h}} - \frac{1}{2h} \right\} =\]
\[\lim_{n \to \infty} \left\{ \frac{1}{1 . 3} + \frac{1}{3 . 5} + \frac{1}{5 . 7} + . . . + \frac{1}{\left( 2n + 1 \right) \left( 2n + 3 \right)} \right\}\]is equal to
\[\lim_{x \to \pi/4} \frac{4\sqrt{2} - \left( \cos x + \sin x \right)^5}{1 - \sin 2x}\] 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
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 limit:
`lim_(x -> 7)[((root(3)(x) - root(3)(7))(root(3)(x) + root(3)(7)))/(x - 7)]`
Evaluate: `lim_(x -> 1) ((1 + x)^6 - 1)/((1 + x)^2 - 1)`
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)]`
Evaluate the following limit:
`lim _ (x -> 5) [(x^3 - 125) / (x^5 - 3125)]`
