Question

If limit of `f(x)=3/2` and `g(x)=1/2` as `x->c`, what the limit of `f(x)/g(x)` as `x->c`?

Answer 1

`3`

Explanation:

`lim_(xrarrc)f(x)/g(x) = (lim_(xrarrc)f(x))/(lim_(xrarrc)g(x))`

provided that `lim_(xrarrc)g(x) != 0`.

We are told that `lim_(xrarrc)g(x) = 1/2`, which is not `0`, so

`lim_(xrarrc)f(x)/g(x) = (lim_(xrarrc)f(x))/(lim_(xrarrc)g(x)) = (3/2)/(1/2)`

`= 3/2 * 2/1 = 3/1 = 3`