lim->

0

sin 3x/7x

i keep getting 0 even using the quotient rule but htis is not the answer

the limit as x approaches 0 of (sinx/x) is 1. Hence, you need to get the given expression, sin3x/7x in the form sinu/u. Observe that sin3x/7x = (sin3x/3x)*(3x/7x)... multiply the numerator and denominator by 3x so we have the expression sinu/u. Thus, the limit as x approaches 0 of (sin3x/3x)*(3x/7x)is simply (1)*(3/7).

You CANNOT use QUOTIENT rule since the question is asking for the VALUE of a particular LIMIT of a function, NOT the slope of the tangent at a particular point. You CAN use L'Hopital's Rule if you know what it is (I suggest you stick with the basics).

To find the limit as x approaches 0 of sin(3x)/(7x), you can use the L'Hospital's Rule, which applies to an indeterminate form of the type 0/0.

First, compute the derivative of sin(3x) and 7x separately. The derivative of sin(3x) is 3cos(3x), and the derivative of 7x is 7. Once you have these derivatives, substitute them back into the original function:

lim(x->0) [3cos(3x)] / 7

Now evaluate the limit as x approaches 0:

lim(x->0) [3cos(3x)] / 7 = (3cos(0)) / 7 = (3 * 1) / 7 = 3/7

Therefore, the limit as x approaches 0 of sin(3x)/(7x) is 3/7, not 0.