lim x--->0 x>2+2x/x?

Answer 1

Hi,
well i think you mean ;
lim (x^2 +2x)/x
x ->0
if you just fill ' 0 ' in to this you will see ,
lim ( 0^2 +2*0)/ 0 = 0/0 and undefined
x ->
so you must simplify this first ;
{ factor ' x ' }
lim x ( x+2) / x { Here ,ignore ' x ' } so we have ;
x-> 0
lim x+2 = 0 + 2 = 2
x -> 0

The result is;
lim (x^2 +2x) / x = +2
x ->0

Good Luck...

Answer 2

Assuming this is,
lim (x^2 +2x) / x
x->0

Just take the x out as it's common from the top.
i.e,
lim x(x+2) / x = lim (x+2) = 2
x->0 x->0

So long as the same term can be cancelled off without the association of a value, just go ahead and cancel it off.

Else use L'Hopital's rule where since the top and bottom are both 0 in the fraction, this ratio is the same as that of the derivatives.
i.e. the limit is the same as
lim (2x+2) / 1 = 2
x->0

Hope this helps!

Answer 3

As stated, you have a false inequality.
Guessing that you meant x^2 instead of x>2,

lim x-->0 x^2+2X/X =2

(Since x never equals 0, the x's in the 2nd term can be canceled, leaving x^2 + 2)

Answer 4

Is this a typo and you really mean

a) lim x-->0 (x^2 + 2x/x)

Or

b) lim x-->0 (x^2 + 2x)/x

Solutions:

a) lim x-->0 (x^2 + 2x/x)
= lim x-->0 (x^2 + 2) (as 2x/x = 2)
= 0^2 + 2
=2

b) lim x-->0 (x^2 + 2x)/x
= lim x-->0 x(x + 2)/x [as x^2 + 2x = x(x+2)]
= lim x-->0 (x + 2) (as x/x = 1)
= 0 + 2
=2

Answer 5

lim x-->0 (x^2 + 2x)/x
= lim x-->0 x(x + 2)/x
[as x-->0 x not equal to 0]
= lim x-->0 (x + 2)
= 0 + 2
=2

Answer 6

answer = a million this is L'well-being center's Rule backwards. L'well-being center's Rule in actuality states that in case you plug the decrease of x (x=0, subsequently) into the equation ([a million-cos^2x]/x^2) and get an indefinite quotient (0/0, ?/?, -?/?, ?/-?,-?/-?), then you particularly can take the spinoff of the numerator and denominator one after the different and plug on your decrease back to get the astonishing answer. (word: L'well-being center's Rule could could be utilized countless situations in case you nonetheless get indefinite quotients, yet it is irrelevant subsequently, in view which you're given the respond, a million) subsequently, L'well-being center's rule could be utilized to the 2nd equation ([a million-cos^2x]/x^2) to get the 1st equation (sinx/x). in view which you're given lim x->0 sinx/x = a million, then you particularly comprehend that lim x->0 (a million-cos^2x)/x^2 = a million. *info* tutor that the 2nd equation will supply an indefinite quotient in case you plug in x=0: [a million-cos^2(0)]/0^2 (a million-a million^2)/0 0/0 (this is an indefinite quotient.) Take the spinoff of the numerator and denominator one after the different: ?(a million-cos^2x)/?x^2 (stick to chain rule) 2sinx/2x (simplify) sinx/x provided that lim x->0 sinx/x = a million, you may state that (a million-cos^2x)/x^2 = a million.

Answer 7

9

Answer 8

I don't know... give up!!