Question

limit x tends to 2 x(x^2-4)/x^-5x+6

Answer 1

S O L U T I O N:

Given Limit:

$\displaystyle \rm =\lim_{x\to2} \dfrac{x(x^{2}-4)}{x^{2}-5x+6}$

If we put x = 2, we get:

$\displaystyle \rm = \dfrac{2\times(4-4)}{2^{2}-5\times2+6}$

$\displaystyle \rm = \dfrac{2\times 0}{4-10+6}$

$\displaystyle \rm = \dfrac{0}{0}$

Which is an indeterminate form.

Lets do some modifications to solve it.

Given Limit:

$\displaystyle \rm =\lim_{x\to2} \dfrac{x(x^{2}-4)}{x^{2}-5x+6}$

Can be written as:

$\displaystyle \rm =\lim_{x\to2} \dfrac{x(x+2)(x-2)}{(x-2)(x-3)}$

$\displaystyle \rm =\lim_{x\to2} \dfrac{x(x+2)}{x-3}$

$\displaystyle \rm =\dfrac{2\times(2+2)}{2-3}$

$\displaystyle \rm =\dfrac{2\times4}{-1}$

$\rm=-8$

Therefore:

$\displaystyle \rm\longrightarrow\lim_{x\to2} \dfrac{x(x^{2}-4)}{x^{2}-5x+6}=-8$

⊕ Which is our required answer.

L E A R N  M O R E:

$\displaystyle \rm1. \: \: \lim_{x \to0} \sin(x) = 0$

$\displaystyle \rm2. \: \: \lim_{x \to0} \cos(x) = 1$

$\displaystyle \rm3. \: \: \lim_{x \to0} \dfrac{ \sin(x) }{x} = 1$

$\displaystyle \rm4. \: \: \lim_{x \to0} \dfrac{ \tan(x) }{x} = 1$

$\displaystyle \rm5. \: \: \lim_{x \to0} \dfrac{1 - \cos(x) }{x} = 1$

$\displaystyle \rm6. \: \: \lim_{x \to0} \dfrac{ \sin^{ - 1} (x) }{x} = 1$

$\displaystyle \rm7. \: \: \lim_{x \to0} \dfrac{ \tan^{ - 1} (x) }{x} = 1$

$\displaystyle \rm8. \: \: \lim_{x \to0} \dfrac{ {e}^{x} - 1 }{x} = 1$

$\displaystyle \rm9. \: \: \lim_{x \to0} \dfrac{ {a}^{x} - 1 }{x} = \ln(a)$

$\displaystyle \rm10. \: \: \lim_{x \to0} \dfrac{ \log(1 + x) }{x} = 1$