Advertisements
Advertisements
प्रश्न
\[\lim_{x \to 2} \frac{x^3 + 3 x^2 - 9x - 2}{x^3 - x - 6}\]
Advertisements
उत्तर
\[\lim_{x \to 2} \left[ \frac{x^3 + 3 x^2 - 9x - 2}{x^3 - x - 6} \right]\] It is of the form \[\frac{0}{0} .\]
Let p(x) = x3 + 3x2 \[-\]9x\[-\] 2
p(2) = 8 + 12 \[-\]18 \[-\]2
= 0
Now,
\[\left( x - 2 \right)\] is a factor of p(x).
Let q(x) = x3 – x – 6
q(2) = 8 – 2 – 6
= 0
\[Now, \left( x - 2 \right)\]is a factor of q(x).
\[\Rightarrow \lim_{x \to 2} \left[ \frac{x^3 + 3 x^2 - 9x - 2}{x^3 - x - 6} \right]\]
\[ = \lim_{x \to 2} \left[ \frac{\left( x - 2 \right)\left( x^2 + 5x + 1 \right)}{\left( x - 2 \right)\left( x^2 + 2x + 3 \right)} \right]\]
\[ = \lim_{x \to 2} \left[ \frac{x^2 + 5x + 1}{x^2 + 2x + 3} \right]\]
\[ = \frac{(2 )^2 + 5 \times 2 + 1}{\left( 2 \right)^2 + 2 \times 2 + 3}\]
\[ = \frac{4 + 10 + 1}{4 + 4 + 3}\]
\[ = \frac{15}{11}\]
APPEARS IN
संबंधित 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?
\[\lim_{x \to 1} \frac{1 + \left( x - 1 \right)^2}{1 + x^2}\]
\[\lim_{x \to 2} \frac{x^4 - 16}{x - 2}\]
\[\lim_{x \to 4} \frac{x^2 - 16}{\sqrt{x} - 2}\]
\[\lim_{x \to 1} \frac{x^3 + 3 x^2 - 6x + 2}{x^3 + 3 x^2 - 3x - 1}\]
If \[\lim_{x \to a} \frac{x^5 - a^5}{x - a} = 405,\]find all possible values of a.
If \[\lim_{x \to a} \frac{x^9 - a^9}{x - a} = \lim_{x \to 5} \left( 4 + x \right),\] find all possible values of a.
\[\lim_{x \to \infty} \frac{\left( 3x - 1 \right) \left( 4x - 2 \right)}{\left( x + 8 \right) \left( x - 1 \right)}\]
\[\lim_{x \to \infty} \sqrt{x + 1} - \sqrt{x}\]
\[\lim_{x \to \infty} \frac{\sqrt{x^2 + a^2} - \sqrt{x^2 + b^2}}{\sqrt{x^2 + c^2} - \sqrt{x^2 + d^2}}\]
\[\lim_{x \to 0} \frac{\sin x \cos x}{3x}\]
\[\lim_{x \to 0} \frac{\sin 5x}{\tan 3x}\]
\[\lim_{x \to 0} \frac{\sin x^n}{x^n}\]
\[\lim_{x \to 0} \frac{\cos ax - \cos bx}{\cos cx - \cos dx}\]
\[\lim_{x \to 0} \frac{1 - \cos mx}{x^2}\]
\[\lim_{x \to 0} \frac{\sin 5x - \sin 3x}{\sin x}\]
\[\lim_{x \to 0} \frac{\tan 3x - 2x}{3x - \sin^2 x}\]
\[\lim_{x \to 0} \frac{x \tan x}{1 - \cos x}\]
\[\lim_{x \to 0} \frac{\cos 2x - 1}{\cos x - 1}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}{x}\]
\[\lim_{x \to 0} \frac{1 - \cos 2x}{3 \tan^2 x}\]
\[\lim_{x \to 0} \frac{\sin 3x + 7x}{4x + \sin 2x}\]
\[\lim_{x \to 0} \frac{\sin ax + bx}{ax + \sin bx}\]
Evaluate the following limit:
\[\lim_{h \to 0} \frac{\left( a + h \right)^2 \sin\left( a + h \right) - a^2 \sin a}{h}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{\cos x} - \sqrt{\sin x}}{x - \frac{\pi}{4}}\]
Evaluate the following limit:
\[\lim_{x \to \pi} \frac{1 - \sin\frac{x}{2}}{\cos\frac{x}{2}\left( \cos\frac{x}{4} - \sin\frac{x}{4} \right)}\]
Write the value of \[\lim_{x \to 0^+} \left[ x \right] .\]
Write the value of \[\lim_{x \to 1^-} x - \left[ x \right] .\]
\[\lim_{x \to \infty} \frac{\sin x}{x} .\]
\[\lim_{n \to \infty} \frac{1^2 + 2^2 + 3^2 + . . . + n^2}{n^3}\]
\[\lim_{x \to \infty} \frac{\sin x}{x}\] equals
\[\lim_{x \to \infty} \frac{\left| x \right|}{x}\] is equal to
Evaluate the following limits: if `lim_(x -> 1)[(x^4 - 1)/(x - 1)] = lim_(x -> "a") [(x^3 - "a"^3)/(x - "a")]`, find all the value of a.
`lim_(x->3) (x^5 - 243)/(x^3 - 27)` = ?
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->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)]`
