Advertisements
Advertisements
प्रश्न
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?
Advertisements
उत्तर
When x < 1, f(x) = a + bx
To find the limit of the left side, keeping the value of x less than 1 and near 1 in f(x),
| x | 0.99 | 0.999 | 0.9999 |
| f(x) | a + 1.99b | a + 0.999b | a + 0.9999b |
∴ `lim_(x → 1^-) f(x) = a + b`
To find the limit of the right-hand side, f(x) = b − ax, with values of x greater than 1 and near 1.
| x | 1.01 | 1.0001 | 1.00001 |
| f(x) | b − 1.01a | b − 1.0001a | b − 1.00001a |
∴ `lim_(x → 1^+) f(x) = b - a`
∴ If limit exists at x = 1 then
`lim_(x → 1^-) f(x) = a + b = b + a = f(1) = 4`
∴ b + a = 4 .....(i)
`lim_(x → 1^+) f(x) = b - a = f(1) = 4`
∴ b − a = 4 ....(ii)
On adding equations (i) and (ii),
2b = 8 or b = 4
Putting b = 4 in equation (i),
4 + a = 4 or a = 0
Hence, a = 0, b = 4
APPEARS IN
संबंधित प्रश्न
\[\lim_{x \to 1} \frac{1 + \left( x - 1 \right)^2}{1 + x^2}\]
\[\lim_{x \to 3} \frac{x^2 - 4x + 3}{x^2 - 2x - 3}\]
\[\lim_{x \to - 1/2} \frac{8 x^3 + 1}{2x + 1}\]
\[\lim_{x \to \sqrt{3}} \frac{x^2 - 3}{x^2 + 3 \sqrt{3}x - 12}\]
\[\lim_{x \to 3} \left( x^2 - 9 \right) \left[ \frac{1}{x + 3} + \frac{1}{x - 3} \right]\]
If \[\lim_{x \to 3} \frac{x^n - 3^n}{x - 3} = 108,\] find the value of n.
If \[\lim_{x \to a} \frac{x^9 - a^9}{x - a} = 9,\] find all possible values of a.
\[\lim_{x \to \infty} \frac{\left( 3x - 1 \right) \left( 4x - 2 \right)}{\left( x + 8 \right) \left( x - 1 \right)}\]
\[\lim_{n \to \infty} \left[ \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + . . . + \frac{1}{3^n} \right]\]
Show that \[\lim_{x \to \infty} \left( \sqrt{x^2 + x + 1} - x \right) \neq \lim_{x \to \infty} \left( \sqrt{x^2 + 1} - x \right)\]
\[\lim_{x \to - \infty} \left( \sqrt{4 x^2 - 7x} + 2x \right)\]
\[\lim_{x \to 0} \frac{\cos 3x - \cos 7x}{x^2}\]
\[\lim_{x \to 0} \frac{\sin^2 4 x^2}{x^4}\]
\[\lim_{x \to 0} \frac{\sin 3x - \sin x}{\sin x}\]
\[\lim_{x \to 0} \frac{\tan 3x - 2x}{3x - \sin^2 x}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}{x}\]
\[\lim_{x \to 0} \frac{\tan 2x - \sin 2x}{x^3}\]
Evaluate the following limit:
\[\lim_{h \to 0} \frac{\left( a + h \right)^2 \sin\left( a + h \right) - a^2 \sin a}{h}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{\cos x} - \sqrt{\sin x}}{x - \frac{\pi}{4}}\]
\[\lim_{x \to \pi} \frac{\sqrt{5 + \cos x} - 2}{\left( \pi - x \right)^2}\]
\[\lim_{x \to 2} \frac{x^2 - x - 2}{x^2 - 2x + \sin \left( x - 2 \right)}\]
\[\lim_{x \to \pi} \frac{\sqrt{2 + \cos x} - 1}{\left( \pi - x \right)^2}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{\cos x - \sin x}{\left( \frac{\pi}{4} - x \right) \left( \cos x + \sin x \right)}\]
Evaluate the following limit:
\[\lim_{x \to \pi} \frac{1 - \sin\frac{x}{2}}{\cos\frac{x}{2}\left( \cos\frac{x}{4} - \sin\frac{x}{4} \right)}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{2 - {cosec}^2 x}{1 - \cot x}\]
\[\lim_{x \to \pi} \frac{\sqrt{2 + \cos x} - 1}{\left( \pi - x \right)^2}\]
\[\lim_{x \to 0} \frac{\sin x^0}{x}\]
\[\lim_{n \to \infty} \left\{ \frac{1}{1 . 3} + \frac{1}{3 . 5} + \frac{1}{5 . 7} + . . . + \frac{1}{\left( 2n + 1 \right) \left( 2n + 3 \right)} \right\}\]is equal to
\[\lim_{x \to 1} \frac{\sin \pi x}{x - 1}\]
If \[\lim_{x \to 1} \frac{x + x^2 + x^3 + . . . + x^n - n}{x - 1} = 5050\] then n equal
\[\lim_{x \to 3} \frac{\sum^n_{r = 1} x^r - \sum^n_{r = 1} 3^r}{x - 3}\]is real to
\[\lim_{n \to \infty} \frac{n!}{\left( n + 1 \right)! + n!}\] is equal to
\[\lim_{x \to \infty} a^x \sin \left( \frac{b}{a^x} \right), a, b > 1\] is equal to
The value of \[\lim_{x \to \pi/2} \left( \sec x - \tan x \right)\]is
\[\lim_{x \to \infty} \frac{\left| x \right|}{x}\] is equal to
Evaluate the following limits: `lim_(x -> "a")[((z + 2)^(3/2) - ("a" + 2)^(3/2))/(z - "a")]`
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)]`
