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

• stunner5000pt
In summary, the first limit does not exist because the numerator goes to 1 while the denominator goes to 0. The second limit, using the substitution method, is correct since u is a variable and does not assume any relationship between x and y.
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?

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.

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.

## 1. What is the definition of a limit in mathematics?

In mathematics, a limit is a fundamental concept that describes the behavior of a function as its input approaches a certain value. It is used to determine the value that a function approaches as its input gets closer and closer to a specified point.

## 2. How do you determine if a limit exists?

A limit exists if the value of the function approaches a single finite number as the input approaches the specified point. This can be determined by evaluating the function at points closer and closer to the specified point and observing if the values approach a single number.

## 3. What is the process for evaluating limits of multivariable functions?

The process for evaluating limits of multivariable functions involves taking the limit along different paths towards the specified point and checking if the values approach the same number. If the values approach the same number along all paths, then the limit exists.

## 4. Is there a shortcut for evaluating limits of multivariable functions?

Yes, there is a shortcut known as the Squeeze Theorem which can be used to evaluate limits of multivariable functions. This theorem states that if two functions approach the same limit as the input approaches a specified point, then any function squeezed between them must also approach the same limit.

## 5. How do you apply the Squeeze Theorem to evaluate the given limit?

To apply the Squeeze Theorem to evaluate the given limit, we need to find two functions that approach the same limit as the input approaches (0,0) and sandwich the given function between them. This will allow us to determine that the limit exists and evaluate its value.

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