Advertisements
Advertisements
Question
If `f(x) = {{:(x + 2",", x ≤ - 1),(cx^2",", x > -1):}`, find 'c' if `lim_(x -> -1) f(x)` exists
Solution
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
RELATED QUESTIONS
\[\lim_{x \to 0} 9\]
\[\lim_{x \to 3} \frac{x^2 - 9}{x + 2}\]
\[\lim_{x \to - 1} \frac{x^3 + 1}{x + 1}\]
\[\lim_{x \to 1} \left( \frac{1}{x - 1} - \frac{2}{x^2 - 1} \right)\]
\[\lim_{x \to 1} \frac{1 - x^{- 1/3}}{1 - x^{- 2/3}}\]
\[\lim_{x \to a} \frac{\left( x + 2 \right)^{3/2} - \left( a + 2 \right)^{3/2}}{x - a}\]
\[\lim_{x \to \infty} \sqrt{x^2 + cx - x}\]
\[\lim_{x \to \infty} \frac{x}{\sqrt{4 x^2 + 1} - 1}\]
\[\lim_{x \to \infty} \left[ \sqrt{x}\left\{ \sqrt{x + 1} - \sqrt{x} \right\} \right]\]
\[\lim_{x \to 0} \frac{\sin 3x}{5x}\]
\[\lim_{x \to 0} \frac{\sin x^0}{x}\]
\[\lim_{x \to 0} \frac{\tan 8x}{\sin 2x}\]
\[\lim_{x \to 0} \frac{\tan mx}{\tan nx}\]
\[\lim_{x \to 0} \frac{\sin 5x}{\tan 3x}\]
\[\lim_{x \to 0} \frac{\cos 3x - \cos 7x}{x^2}\]
\[\lim_{x \to 0} \frac{x \cos x + 2 \sin x}{x^2 + \tan x}\]
\[\lim_{x \to 0} \frac{1 - \cos 2x + \tan^2 x}{x \sin x}\]
\[\lim_{x \to 0} \frac{x \tan x}{1 - \cos 2x}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}{x}\]
\[\lim_{x \to 0} \frac{1 - \cos 5x}{1 - \cos 6x}\]
\[\lim_{x \to 0} \frac{\sin ax + bx}{ax + \sin bx}\]
\[\lim_{x \to \frac{\pi}{2}} \frac{1 - \sin x}{\left( \frac{\pi}{2} - x \right)^2}\]
\[\lim_{x \to \frac{3\pi}{2}} \frac{1 + {cosec}^3 x}{\cot^2 x}\]
\[\lim_{x \to 0} \frac{\log \left( a + x \right) - \log a}{x}\]
\[\lim_{x \to \infty} \frac{\sin x}{x} .\]
Write the value of \[\lim_{x \to 0} \frac{\sin x^\circ}{x} .\]
\[\lim_{x \to 3} \frac{x - 3}{\left| x - 3 \right|},\] is equal to
If `lim_(x -> 1) (x^4 - 1)/(x - 1) = lim_(x -> k) (x^3 - l^3)/(x^2 - k^2)`, then find the value of k.
Evaluate the following limit :
`lim_(x->7)[[(root3(x)- root3(7))(root3(x) + root3(7)))/(x-7)]`
Evaluate the following limit.
`lim_(x->3)[sqrt(x + 6)/x]`
