I got this calculas question in an exam and couldn't answer it, could appreciate some help.

Question

Limit as x tends to 0

x²+Sin3x ÷ 5xCosx

Answer 1

x²+Sin3x ÷ 5xCosx

The limit as x ---> 0 is indeterminate, so we would need to try L'Hopital's rule. Take the first derivative of top and bottom to get:

2x + 3cos(3x) / -5xsinx + 5cosx

As x ---> 0, this becomes: 3cos(3x) / 5cosx = 3/5

CHECK by plotting the original function using a graphing calculator or online tool

Answer 2

Another way to evaluate the limit without resorting to L'Hospital's Rule is to use the fact that

lim_θ→0 {sinθ / θ) = 1.

So, lim_x→0 [ (x² + sin(3x)) / (5xcosx)]

= lim_x→0 [ (x + sin(3x)/x) / (5cosx) ]

= lim_x→0 [ (x + 3sin(3x) / (3x)) / (5cosx) ] = (0 + 3(1)) / (5(1)) = 3/5

Answer 3

x²+Sin3x ÷ 5xCosx

The limit as x ---> 0 is indeterminate, so we would need to try L'Hopital's rule. Take the first derivative of top and bottom to get:

2x + 3cos(3x) / -5xsinx + 5cosx

As x ---> 0, this becomes: 3cos(3x) / 5cosx = 3/5

CHECK by plotting the original function using a graphing calculator or online tool