Advertisements
Advertisements
Question
\[\lim_{x \to 5} \frac{x^3 - 125}{x^2 - 7x + 10}\]
Solution
\[\lim_{x \to 5} \left[ \frac{x^3 - 125}{x^2 - 7x + 10} \right]\]
\[\text{ It is of the form } \frac{0}{0} . \]
\[ \lim_{x \to 5} \left[ \frac{x^3 - 5^3}{x^2 - 2x - 5x + 10} \right]\]
\[ = \lim_{x \to 5} \left[ \frac{\left( x - 5 \right)\left( x^2 + 5x + 5^2 \right)}{x\left( x - 2 \right) - 5\left( x - 2 \right)} \right] \left[ \because A^3 - B^3 = \left( A - B \right)\left( A^2 + AB + B^2 \right) \right]\]
\[ = \lim_{x \to 5} \left[ \frac{\left( x - 5 \right)\left( x^2 + 5x + 25 \right)}{\left( x - 2 \right)\left( x - 5 \right)} \right]\]
\[ = \frac{5^2 + 5 \times 5 + 25}{5 - 2}\]
\[ = \frac{75}{3}\]
\[ = 25\]
APPEARS IN
RELATED QUESTIONS
\[\lim_{x \to - 1}{\left( 4 x^2 + 2 \right)}\]
\[\lim_{x \to 0} \frac{3x + 1}{x + 3}\]
\[\lim_{x \to - 5} \frac{2 x^2 + 9x - 5}{x + 5}\]
\[\lim_{x \to 3} \left( \frac{1}{x - 3} - \frac{3}{x^2 - 3x} \right)\]
\[\lim_{x \to 3} \left( x^2 - 9 \right) \left[ \frac{1}{x + 3} + \frac{1}{x - 3} \right]\]
\[\lim_{x \to 27} \frac{\left( x^{1/3} + 3 \right) \left( x^{1/3} - 3 \right)}{x - 27}\]
\[\lim_{x \to \infty} \left[ \left\{ \sqrt{x + 1} - \sqrt{x} \right\} \sqrt{x + 2} \right]\]
\[\lim_{x \to \infty} \left[ \frac{x^4 + 7 x^3 + 46x + a}{x^4 + 6} \right]\] where a is a non-zero real number.
\[\lim_{x \to 0} \frac{x^2}{\sin x^2}\]
\[\lim_{x \to 0} \frac{\sin 5x}{\tan 3x}\]
\[\lim_{x \to 0} \frac{\cos 3x - \cos 7x}{x^2}\]
\[\lim_{x \to 0} \frac{x \cos x + 2 \sin x}{x^2 + \tan x}\]
\[\lim_{x \to 0} \frac{5 x \cos x + 3 \sin x}{3 x^2 + \tan x}\]
\[\lim_{x \to 0} \frac{\sec 5x - \sec 3x}{\sec 3x - \sec x}\]
\[\lim_{x \to 0} \frac{x^2 - \tan 2x}{\tan x}\]
\[\lim_{x \to 0} \frac{x \tan x}{1 - \cos x}\]
Evaluate the following limit:
\[\lim_{h \to 0} \frac{\left( a + h \right)^2 \sin\left( a + h \right) - a^2 \sin a}{h}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{1 - \tan x}{x - \frac{\pi}{4}}\]
\[\lim_{x \to \frac{\pi}{8}} \frac{\cot 4x - \cos 4x}{\left( \pi - 8x \right)^3}\]
\[\lim_{x \to 2} \frac{x^2 - x - 2}{x^2 - 2x + \sin \left( x - 2 \right)}\]
\[\lim_{x \to \frac{\pi}{2}} \left( \frac{\pi}{2} - x \right) \tan x\]
\[\lim_{x \to \frac{\pi}{2}} \frac{\left( \frac{\pi}{2} - x \right) \sin x - 2 \cos x}{\left( \frac{\pi}{2} - x \right) + \cot x}\]
\[\lim_{x \to 0} \frac{\sin 2x}{e^x - 1}\]
\[\lim_{x \to 0} \frac{8^x - 2^x}{x}\]
Write the value of \[\lim_{x \to 1^-} x - \left[ x \right] .\]
Write the value of \[\lim_{x \to 0} \frac{\sin x^\circ}{x} .\]
\[\lim_{n \to \infty} \frac{1^2 + 2^2 + 3^2 + . . . + n^2}{n^3}\]
\[\lim_{h \to 0} \left\{ \frac{1}{h\sqrt[3]{8 + h}} - \frac{1}{2h} \right\} =\]
\[\lim_{n \to \infty} \left\{ \frac{1}{1 . 3} + \frac{1}{3 . 5} + \frac{1}{5 . 7} + . . . + \frac{1}{\left( 2n + 1 \right) \left( 2n + 3 \right)} \right\}\]is equal to
\[\lim_{x \to 1} \frac{\sin \pi x}{x - 1}\]
\[\lim_{x \to \pi/4} \frac{4\sqrt{2} - \left( \cos x + \sin x \right)^5}{1 - \sin 2x}\] is equal to
The value of \[\lim_{x \to \pi/2} \left( \sec x - \tan x \right)\]is
The value of \[\lim_{n \to \infty} \left\{ \frac{1 + 2 + 3 + . . . + n}{n + 2} - \frac{n}{2} \right\}\]
Evaluate the following limits: `lim_(x -> 0)[((1 - x)^8 - 1)/((1 - x)^2 - 1)]`
Let `f(x) = {{:((k cos x)/(pi - 2x)",", "when" x ≠ pi/2),(3",", x = pi/2 "and if" f(x) = f(pi/2)):}` find the value of k.
Number of values of x where the function
f(x) = `{{:((tanxlog(x - 2))/(x^2 - 4x + 3); x∈(2, 4) - {3, π}),(1/6tanx; x = 3"," π):}`
is discontinuous, is ______.
If f(x) = `{{:(1 if x "is rational"),(-1 if x "is rational"):}` is continuous on ______.
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->3)[sqrt(x + 6)/x]`
