Question

What is the limit as x goes to zero for:
((sin(2x))/(7x)?

Answer 1

pls explain any answer - thanks a ton!@

Answer 2

Remember the special limit for sin(x)/x ?

Answer 3

=1

Answer 4

The easiest way to do this problem is to use L'Hopital's rule.
If you plug in 0 into the fraction you get 0/0, which is the parameter for using L'Hopital's.
So the rule goes, if top and bottom equal zero, you can take the derivative of top and bottom until one does not go to zero.
\[d(\sin(2x))=2\cos2x\]
\[d(7x)=7\]
so you get \[2\cos(2x)/7\]
where x goes to zero, so plug it in and you get 2/7.
The limit is 2/7

Answer 5

Good. It works the same for sin(nx)/nx were n is some number.

Answer 6

cool, hope my explanation wasn't too complicated

Answer 7

one sec< while I process - thanks all> I have abt 5 of these q's and I really want to get them

Answer 8

Have you used L'Hopitals Rule yet?

Answer 9

True, if you haven't dealt with L'Hopital's yet then you should go with cruffo's explanation

Answer 10

yes, i think so

Answer 11

how does the derivative of sin(2x) become 2cos(2x)?

Answer 12

This problem is a typical "first time using limits" problem, in which case you would go with an algebraic solution, since you haven't covered derivative rule yet.

Answer 13

that has really confused me :-) sorry

Answer 14

we just covered the pinching theorem

Answer 15

Ok, put PJW_Largo's solution aside for about 3 weeks :)

Answer 16

yeah sorry about that
you'll probably get to those derivatives soon though

Answer 17

no worries; we pull out the number next to the x ( not by the trig function)

Answer 18

Try this,
\[\frac{sin(2x)}{7x} \]
\[= \frac{1}{7} \frac{\sin(2x)}{x}\]
\[ = \frac{1}{7}\cdot\frac{2}{2}\cdot\frac{\sin(2x)}{x}\]

Answer 19

forget l'hopital because you have not got there yet. you know
\[lim_{x->0}\frac{sin(x)}{x}=1\] so
\[lim_{x->0} \frac{sin(2x)}{7x}=\frac{2}{7}\]

Answer 20

yay! I get it!!!!

Answer 21

rock n roll!!!

Answer 22

ok, 5 more here we go!

Answer 23

:)