Question

find the limit of the function

limit sin3x/8x
x->0

Answer 1

thanks so much

Answer 2

No problemo! Have fun!
use l'hopital sin3x turns into 3(cosx) and 8x derives into 8 so, the answer is 3/8

Do let me know if you don't understand it. I'd love to explain it since I just learned this this morning. Hee hee.

Answer 3

i wanted to ask how does the sin3x turn into 3(cosx)?

Answer 4

Sure thing.

The derivative of sinx is always cosx, right? (It's one of the 'formulas' you should memorize).

The 3 in front of X may be moved in front of the sin, so you have 3sinX. After using l'hopital's rule, you get 3(cosx).

Answer 5

The derivative of sin(3x) is NOT 3cos(x) its 3cos(3x).

Answer 6

okay thanks

Answer 7

\[\frac{3}{8}\] with your eyeballs

Answer 8

you can solve this using either L hospital's rule or in simple way:-
to apply L hospital rule it is necessary to check that function should give 0/0or (infinity/infinity) when we give x=0
here when we put x=0 in sin3x/8x it gives 0/0 hence we can apply L hospital's rule;
what rule says is just differentiate numerator and denominator until you get the constant value;
-> (d/dx (sin3x) ) / (d/dx of 8x)
-> 3cos3x /8 ...now see your denominator is free from x i.e if you put x=0 now your denominator can't give 0;
put x=0 above
-> 3 cos0 /8 =3/8 answer

Answer 9

you need no l'hopital or anything of that sort. you can see it clearly, so forget l'hopital

Answer 10

lhopital is so much easier than anything else, though!

Answer 11

only one thing you need to know, namely that
\[lim_{x\rightarrow 0} \frac{\sin(x)}{x}=1\]

Answer 12

@uber
yes but that is like saying that it is so much easier to kill a mouse with an M16

Answer 13

if you have already learned lots about derivatives etc. then you can apply l'hopital. but typically these questions arrive in a calc text before even derivatives. so there is no using l'hopital.

Answer 14

@galactic what are you presumed to know before you can answer this question?

Answer 15

that sin(x) = -cos(x) that i remember we learned in class the other day

Answer 16

That's not a great example, though.
Your example is using a difficult method compared to my uber-easy method.

It's more like I'm using a grenade to kill a mouse.

Using a M16 is more like using the difference quotient vs using your method. Mine is FARRRR easier than any other method.

Answer 17

cos(x) = -sin(x) that's what i meant

Answer 18

Oh phew. Scared me there for a moment. :P

Though: You mean to say "the DERIVATIVE of cosx is -sinx"

Because cosx doesn't equal -sinx. Make sure you know the difference. ^_^

Answer 19

hahah yeah i confused myself for a minute....

Answer 20

if you have not yet taken the derivative of anything, let alone used l'hopital (Hospital) rule you simply do this: you know that
\[lim_{x\rightarrow 0}\frac{\sin(x)}{x}=1\] so of course
\[lim_{x\rightarrow 0}\frac{\sin(3x)}{3x}=1\] and so
\[lim_{x\rightarrow 0}\frac{8\sin(3x)}{3x}=8\] therefore
\[lim_{x\rightarrow 0} \frac{\sin(3x)}{8x}=\frac{3}{8}\]

Answer 21

yea i haven't use the l'hopital yet so the examples you put really help understand

Answer 22

that is what i assumed. hope the method is both easy and clear.

Answer 23

yeah it is...the trig functions are sometimes tricky so i take time to understand those one more in depth

Answer 24

with a small (very small) amount of practice you do it with your eyeballs
\[lim_{x\rightarrow 0}\frac{\sin(ax)}{b}=\frac{a}{b}\]

Answer 25

typo i meant
\[lim_{x\rightarrow 0}\frac{\sin(ax)}{bx}=\frac{a}{b}\]

Answer 26

haha its okay..and yea the book problems help alot so my understanding of this stuff is better
though a month of calculus is a bit complex the work keeps me active

Answer 27

mind wise*