Does this limit exist lim e^(r^2)/(cos(t)sin(t)) for (r,t)->(0,0)

[stunner5000pt]

In my exam the question was to determine the existence of this limit
\lim_{(r,t)\rightarrow(0,0)} \frac{e^{r^2}}{\cos{t}\sin{t}}

now i wrote the numerator has no t associated with it so grows or shrinks without bounds, the same applies for the denominator...
so the limit does not exist

is this a good reason

another question was
\lim_{(x,y)\rightarrow(0,0)} \frac{\sin{xy}}{xy}
i substituted xy = u and got \lim_{u \rightarrow 0} \frac{\sin{u}}{u} = 1


is this the correct method?
Am i right?

 

[HallsofIvy]

The numerator does not "grow or shrink without bound". As r goes to 0, the numerator goes to 1.

However, you are correct that the denominator goes to 0 as t does no matter what r is. Since the numerator goes to 1, what does that tell you about the fraction?

I was a bit suspicious about your second method, but yes, it works, since u is a variable, you are not assuming any relationship between x and y.

 

[thepatientmental]

Your reasoning for the first limit is not entirely correct. While it is true that the numerator and denominator both grow without bounds as (r,t) approaches (0,0), this does not necessarily mean that the limit does not exist. In order to determine the existence of a limit, you need to consider the behavior of the function as a whole, not just the individual components. In this case, as (r,t) approaches (0,0), the function becomes oscillatory, with the numerator and denominator both approaching 0. This type of behavior can lead to the existence of a limit. However, in this case, the function is not well-defined at (0,0) since the denominator becomes 0, so the limit does not exist.

Your approach for the second limit is correct. By substituting xy=u, you are essentially reducing the problem to a single variable limit, which can be evaluated using standard techniques. In this case, the limit evaluates to 1, so you are correct in saying that the limit exists and is equal to 1.