Question

State the value of the limit, if it exists

lim x→0 sin(x) multiplied with 3x^3+2x^2/x^2

Anybody? :)

Answer 1

ok this is a big one

Answer 2

so stay with me

Answer 3

first we separate the lim

Answer 4

\[\lim_{x \rightarrow 0} (3x+2) \times \lim_{x \rightarrow 0}(\frac{ \sin(x) }{ x ^{2} })\]

Answer 5

i forgot to times x^2 with sin(x)

Answer 6

ok

Answer 7

\[\lim_{x \rightarrow 0}(\frac{ \sin(x)x ^{2} }{ x ^{2} }) \times \lim_{x \rightarrow 0} (3x+2)\]

Answer 8

ur following the steps right

Answer 9

wow! :)

Answer 10

Yep! Think so…

Answer 11

3x+2 is a polynomial and thus everywhere continuous so \[\lim_{x \rightarrow 0} (3x+2)= 2+3(0)=2\]

Answer 12

and the \[\lim_{x \rightarrow 0} \frac{ \sin(0) \times 0 }{ 0 } =0\]

Answer 13

\[ 2 \times 0=0\]

Answer 14

got it

Answer 15

if u have similar denominator like x^2 in ur case u can separate the limit

Answer 16

Yup! Think so…. :)

Answer 17

I have to go through it a few times to make sure I understand everything :)

Answer 18

Thank you so much!

Answer 19

Any possibilities of some follow-up questions later on?

Answer 20

Yea sure

Answer 21

you simply can not say this:
\[\lim_{x \rightarrow 0}(\frac{ \sin(x)x ^{2} }{ x ^{2} }) \]
\[\implies \lim_{x \rightarrow 0} \frac{ \sin(0) \times 0 }{ 0 } =0\]

the whole point of this is that dividing by zero you cannot do, so you check what the function in question does as you approach very close to zero. so:
\[\lim\limits_{ x→0} \, \, \sin(x) .\frac{3x^3+2x^2}{x^2}\]
\[=\lim\limits_{ x→0} \, \, \sin(x) .\lim\limits_{ x→0}\frac{3x^3+2x^2}{x^2}\]
\[=\lim\limits_{ x→0} \, \, \sin(x) .\lim\limits_{ x→0}\frac{\frac{3x^3}{x^2}+\frac{2x^2}{x^2}}{\frac{x^2}{x^2}}\]
\[=\lim\limits_{ x→0} \, \, \sin(x) .\lim\limits_{ x→0}\frac{3x+2}{1}\]
\[=0 \times 2 = 0\]

Answer 22

Thank you so much for your reply! :)