Find value to make function continuous

Find a such that f(x) = 9x / tanx, x does not equal 0
= a if x = 0
is continuous at x=0

9x/tanx
F1' 9x = 9
F2'; tan(x) = sec(x)^2

F1(0) =0
F2(0) = 0.

indeterminate case so we can use
Hospital's rule


lim x goes to 0 of 9x/tan(x) = lim x-0 9/sec(0)^2
sec(x) = 1/cos(x) cos (0) = 1 so sec(0) = 1 so sec^2(0) = 1

F1'(0) = 9.
F2'(0) = 1.

so the limit at 0 is =9/1 = 9

Limit is 9, value is 0, it's not
continuous at zero

To do this, we must find:

lim x->0 of 9x/tan(x). Plugging in 0 gives 0/0, indeterminate form, so we can use L'Hopital's rule:

lim x->0 of 9/sec^2(x) is the resulting problem, and when plugging in 0 for (x), we get 9 / 1, or 9.

So, if a = 9, the function is continuous at x = 0.