Question

need help on the attachment

Answer 1

\[
\lim_{x\rightarrow 0} \left(1+\frac{\sin(-4x)}{3x}\right)
= \lim_{x\rightarrow 0}\left(1+\frac{-4x}{3x}\right)=1-\frac{4}{3}=-\frac{1}{3}
\]
The first equality holds because of the Taylor expansion of sin. Only the linear term needs to be considered, because all the higher terms vanish as x->0.

Answer 2

couldn't you use the quotient rule?

Answer 3

You mean l'Hopital? Yes you could. That gives you the same answer.

Answer 4

because my professor haven't taught us the "Taylor" expansion of sin. but if you use the quot2ient rule wouldn't you get -3x*(Cos x)+(4sinx)(3)/(3x)^2

Answer 5

is that right

Answer 6

Not quite.

Answer 7

It's not the quotient rule that you use here! It's l'Hopital.

Answer 8

\[ \lim_{x\rightarrow 0} \frac{f(x)}{g(x)} = \lim_{x\rightarrow 0} \frac{f^\prime(x)}{g^\prime(x)} \]

Answer 9

Apply that to the quotient inside the limit!

Answer 10

how come you didn't do anything to the 1 inside the limit?

Answer 11

The 1 contains no x.

Answer 12

So it just comes out of the limit unchanged.

Answer 13

No apply the formula above to f(x)=sin(-4x) and g(x)=3x

Answer 14

-4*cosx/3

Answer 15

-4*cos4x/3

Answer 16

No you can plug in x=0, as there is no x in the denominator left.

Answer 17

*Now

Answer 18

wait negative stay inside the -4cos(-4x), or -4cos(4x)?

Answer 19

inside the paranthesis

Answer 20

Doesn't matter a) because cos(-x)=cos(x) and b) because you set x=0 anyway

Answer 21

Both what if it did matter which way would be correct?

Answer 22

They are both correct because cos is an even function!

Answer 23

cos(x) = cos(-x) for all x

Answer 24

ok

Answer 25

-4Cos(4*0)=-4*1=-4

Answer 26

Yes.

Answer 27

answer -1/3, so l'Hopital can always be used if you have scenarios like this one?

Answer 28

Its always useful if you have to evaluate the limit of a quotient of differentiable functions.

Answer 29

ok, thanks for the help.