Question

lim(x->0) [(3x+4X^2)/(x^2-x^3)] = ?

Wolfram alpha says this limit does not exist.

Could anyone check where am I wrong?

(3x+4x^2)/(x^2-x^3)

= (3+4x)/(x-x^2)

= 3/(x-x^2) + 4x/(x(1-x))

=> lim(x->0)[3/(x-x^2) ]+lim(x->0)[4x/(x(1-x))]

plug zero in

+infinite + 4 = +infinite

Answer 1

\[\lim_{x\rightarrow 0}\frac{3x + 4x^2}{x^2 - x^3}\]

Answer 2

Start calculating derivatives of numerator and Denominator to simplify expression until there is no x in denominator.. and then put x = 0
if this isnt possible , then limit doesnt exist :)

Answer 3

Use the power rule of differetiation : d/dx (x^y) = yx^y-1

Answer 4

I'm not allowed to use derivative =)

Answer 5

lol

Answer 6

It's the simplest way there..
Effort time ! :)

Answer 7

yeah, the CAS is saying that the limit does not exist.

try doing l'hospital

\[\lim_{x\rightarrow 0}\frac{3 + 8x}{2x - 3x^2} = \infty\]

Answer 8

Here , degree of denominator is higher than that of numerator.. So you can specify this reason to say "Limit does not exist"

Answer 9

right.

Answer 10

oh i see it's also an odd number thanks.

Answer 11

vik, are you sure that's enough to say so? Because I was thinking.

lim x-> 0 1/x^2

Denominator is higher grade than numerator, but the limit exists, which is plus infinity.
The reason why the limit doesn't exist, according to me, is the same reason why

lim x->0 |x|/x doesn't exist. Doing lim->0+ and lim->0- we find different values.