Question

limit sin10x-2x/3x
x-->0

Answer 1

\[\lim_{x \rightarrow 0}\frac{\sin(10x)-2x}{3x}\]is this correct?

Answer 2

yeah

Answer 3

l'Hospital's rule and plug in x=0

Answer 4

am a novice...we just started with limits on calculus..

Answer 5

havent reached that level

Answer 6

I'm not sure how to do it without l'Hospital's rule, which is to take the derivative of the top and bottom when you get 0/0 by plugging in:\[\lim_{x \rightarrow 0}\frac{\sin(10x)-2x}{3x}=\lim_{x \rightarrow 0}\frac{10\cos(10x)-2}{3}=\frac{8}{3}\]Feel free not to give me a medal if you want to wait for someone to do it without l'Hospital.

Answer 7

what about u = 10x into
\[\lim_{u->0} (\frac{\sin(u)}{\frac{3}{10}u}-\frac{2}{3}) = \frac{10}{3}-\frac{2}{3} = 0\]

however you're not getting anywhere without \[\lim_{u->0} \frac{\sin(u)}{u} = 1\]

Answer 8

oops i mean 8/3

Answer 9

^still, probably more what the asker was looking for, even taking
lim x->0(sinx)/x for granted.
Are you a student Broken?
You seem pretty sharp.

Answer 10

\[\lim_{u->0} (\frac{\sin(u)}{\frac{3}{10}u}-\frac{2}{3}) = \lim_{u->0} \frac{10}{3} \frac{\sin(u)}{u}-\frac{2}{3} =\frac{10}{3}-\frac{2}{3} = \frac{8}{3}\]
finished studenting and moved on to helping students

Answer 11

we have already given you the answer in two ways krypton. Find your mistake, it's 8/3

Answer 12

ok,,thanks dear