Question

lim x->0 sin3x/5x^3-4x.

Answer 1

answer is -3/4

Answer 2

I want to know how to get to the answer -3/4

Answer 3

Have you learnt the l'Hôpital's rule yet?

Answer 4

no, just the trig identities/chain rule. We are not allowed to use the l'hopital's rule yet.

Answer 5

Chain rule? for limit? or differentiation?

Answer 6

well for this part we suppose to simplify it to solve it.

Like, the previous question was lim x->0 sin4x/sin6x, so we times (4/6) to both top and botom, making it sin4x/sin4x since we bring out the 4/6 to the front, so it's 4/6 lim x->0 sin 4x/4x (and let 4x= theta)=> 4/6 times 1=> 4/6 => 2/3. Just a sample of what we're suppose to be doing.

Answer 7

Got it. Here it goes.
\[\lim_{x \rightarrow 0}\frac{ \sin3x }{ 5x^3-4x }\]\[=\lim_{x \rightarrow 0}\frac{ \sin3x }{3x}\times \frac{3x}{ 5x^3-4x }\]\[=\lim_{x \rightarrow 0}\frac{ \sin3x }{3x}\times \lim_{x \rightarrow 0}\frac{3x}{ x(5x^2-4) }\]\[=\lim_{x \rightarrow 0}\frac{ \sin3x }{3x}\times \lim_{x \rightarrow 0}\frac{3}{ 5x^2-4 }\]Can you continue? And do you understand what I've done?

Answer 8

Btw, I think \[\lim_{x \rightarrow 0}\frac{ \sin x }{x}=1\]Amd\[\lim_{x \rightarrow 0}\frac{ \sin 3x }{3x}=\lim_{u \rightarrow 0}\frac{ \sin u }{u} = 1\]

Answer 9

How'd you jump from sin 3x/5x^3-4x => sin3x/3x in the first step?

Answer 10

Not really jump...
Note that \[\lim_{x \rightarrow 0}\frac{ \sin 3x }{3x}= 1\]Now, we have \[\lim_{x \rightarrow 0}\sin 3x \], and we have to use \[\lim_{x \rightarrow 0}\frac{ \sin 3x }{3x}= 1\] tp evaluate the limit, so, multiply the limit by 3x/3x, which is 1.
\[\lim_{x \rightarrow 0} (\sin 3x \times \frac{3x}{3x}) = \lim_{x \rightarrow 0}(\frac{ \sin 3x }{3x}\times 3x)\]

Answer 11

*to evaluate

Answer 12

but doesn't that just bring it back to the original equation ?

Answer 13

Because sin3x is not the original question and you're just asking me a part of it... Here it goes again:
\[\lim_{x \rightarrow 0}\frac{ \sin3x }{ 5x^3-4x }\]\[\lim_{x \rightarrow 0}(\sin3x \times \frac{ 1 }{ 5x^3-4x })\]\[=\lim_{x \rightarrow 0}(\sin3x\times \frac{3x }{3x}\times \frac{1}{ 5x^3-4x })\]\[=\lim_{x \rightarrow 0}\frac{ \sin3x }{3x}\times \frac{3x}{ 5x^3-4x }\]\[=\lim_{x \rightarrow 0}\frac{ \sin3x }{3x}\times \lim_{x \rightarrow 0}\frac{3x}{ x(5x^2-4) }\]\[=\lim_{x \rightarrow 0}\frac{ \sin3x }{3x}\times \lim_{x \rightarrow 0}\frac{3}{ (5x^2-4) }\]\[=...\]

Answer 14

Ok I think I got it. Thanks

Answer 15

Welcome :)