Someone please walk me through this calculus limit problem?

Question

lim (x approaches 0) (sin3t)/(2t)

Explain step by step please.

Answer 1

If you have learnt that lim(sinx / x) = 1 when x--> 0,  then you can find this limit easily.

lim (sin3t)/(2t) = lim [(sin3t) /(3t)(2/3)] = (3/2) lim(sin3t)/(3t)

t-->0

As t approaches 0, 3t also approaches 0, therefore

lim (sin3t)/(3t) = 1

t-->0

lim (sin3t)/(2t) = (3/2)*1 = 3/2

t-->0

Answer 2

Plug 0 in for t. sin(3(0))/2(0) = 0/0

Use l'Hôpital's rule by taking the derivative of both the numerator and the denominator and plug 0 in again for t.  sin(3(t))/2(t) = 3 cos 3(0)/2 = 3/2 = ANSWR.

Answer 3

Notice that if we try to evaluate the limit by direct substitution we get sin(3*0)/(2*0)=0/0=indeterminate

So we apply l'hopital's rule. Differentiate the numerator and denominator separately then evaluate the limit.

lim f(x)=3sin3t/2=(3cos(3*0))/2=3/2

x--->0

limit is 3/2.

Answer 4

*EDIT* Please ignore what I wrote before. I completely forgot about L'Hospital's Rule. So let me redo this :P .

L'Hospital's Rule says that if direct substitution for a limit leads to an indeterminate form (i.e. 0/0, ∞^0,∞-∞, etc.), you can use the derivatives of the numerator and denominator to evaluate the limit.

In this case, direct substitution of t=0 leads to 0/0, so this rule applies.

We take the derivative of the top:

d/dx[sin(3t)]=3cos(3t)

And the derivative of the bottom:

d/dt[2t]=2

So we now have this:

lim(x->0) [3cos(3t)/2]

Since this is a continuous function for all t, direct substitution yields:

3cos(0)/2=3/2

So it's 3/2.

-IMP ;) :)

Answer 5

I assume you mean lim (t approaches 0) (sin3t)/(2t):

lim (t approaches 0) (sin3t)/(2t)

= lim (t approaches 0) (3/2) (2/3) (sin3t)/(2t)

= (3/2) lim (t approaches 0) (2/3) (sin3t)/(2t)

= (3/2) lim (t approaches 0) (sin3t)/(3t)

= (3/2) lim (u approaches 0) (sin u)/(u)

= (3/2) 1

= 3/2

(Above I used a substitution of u for 3t and the fact that

lim (u approaches 0) (sin u)/(u) = 1).

You can also use L'Hopital's rule and take the derivative of numerator and denominator and set t = 0 -- it provides the same answer.