Notice we cannot just substitute 0 because `(sin\ 0)/0` is undefined.
There is no algebraic process to find this limit. We can substitute values of x which get closer and closer to `0` (from both the left side and right side) and conclude that
A way to check this is to graph it and see that indeed the limit as x gets closer to `0` is `1`:
We cannot see it, but there is a "hole" at `x = 0` in our graph.
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