Try it on paper first. A hint, the key idea and the worked solution are here when you need them.
Let wherever it is defined. It is not defined at two points. At one of them the break is removable, and at the other it is infinite. Find the limit of at the removable point, and the number of infinite breaks.
On an interval, check every joining point of a piecewise function: the left-hand limit, the right-hand limit and the value must be equal. Sums, products, quotients (with a nonzero denominator) and compositions of continuous functions are continuous.
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Step 1. Factorise. . It is undefined at and at .
Step 2. The point . For , . As this tends to . The limit exists, so this break is removable.
Step 3. The point . Near the simplified form has top and a bottom that tends to zero. The values grow without bound, so this break is infinite.
Answer: the limit is , and there is infinite break.
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