Question

lim x tends to 0 5x³+8x²/3x⁴-16x²​

Answer 1

S O L U T I O N:

Given Limit:

$\displaystyle\rm=\lim_{x\to0}\dfrac{5x^{3}+8x^{2}}{3x^{4}-16x^{2}}$

Can be written as:

$\displaystyle\rm=\lim_{x\to0}\dfrac{x^{2}(5x+8)}{x^{2}(3x^{2}-16)}$

$\displaystyle\rm=\lim_{x\to0}\dfrac{5x+8}{3x^{2}-16}$

Now put x = 0, we get:

$\displaystyle\rm=\dfrac{5\times0+8}{3\times0^{2}-16}$

$\displaystyle\rm=\dfrac{8}{-16}$

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

$\rm=-2^{-1}$

Therefore:

$\displaystyle \rm\longrightarrow\lim_{x\to0}\dfrac{5x^{3}+8x^{2}}{3x^{4}-16x^{2}}=-2^{-1}$

⊕ Which is our required answer.

L E A R N  M O R E:

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

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

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

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

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

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

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

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

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

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