Question

Evaluate the following limits. $$\lim _{(x, y) \rightarrow(0,2)}(2 x y)^{x y}$$

Short answer

Question: Determine the value of the limit as (x, y) approaches (0, 2) for the function \((2xy)^{xy}\). Answer: The limit of the given function as (x, y) approaches (0, 2) is 1.

Step-by-step solution

Write the given function in the exponential form

We have the function \((2xy)^{xy}\), which can be rewritten as \(e^{xy \ln(2xy)}\) using the property \(a^b = e^{b\ln(a)}\).

Apply substitution

Let \(t = xy\). Then, when \((x, y) \rightarrow (0, 2)\), \(t \rightarrow 0\). So, now consider the function \(e^{t\ln(2t)}\), and we will now find the limit as \(t \rightarrow 0\) instead.

Rewrite the limit

We have now the function \(e^{t\ln(2t)}\). We need to find \(\lim_{t \rightarrow 0} e^{t\ln(2t)}\). Using properties of limits, we can rewrite the function as \(e^{\lim_{t \rightarrow 0} (t\ln(2t))}\).

Simplify the limit

Now we must find the limit of the exponent \(\lim_{t \rightarrow 0} (t\ln(2t))\). This is an indeterminate form of type \(0\cdot\infty\). To simplify, we apply L'Hôpital's rule to the equivalent expression \(\frac{\ln(2t)}{\frac{1}{t}}\).

Apply L'Hôpital's Rule

Using L'Hôpital's Rule, we differentiate both the numerator and denominator. \(\frac{d(\ln(2t))}{dt} = \frac{1}{t}\) and \(\frac{d(\frac{1}{t})}{dt} = -\frac{1}{t^2}\). The limit becomes \(\lim_{t \rightarrow 0} \left(\frac{\frac{1}{t}}{-\frac{1}{t^2}}\right)\), which can be rewritten as \(\lim_{t \rightarrow 0} (-t)\).

Evaluate the limit

The limit \(\lim_{t \rightarrow 0} (-t) = 0\).

Find the final answer

We now know that the limit in the exponent is \(0\). The final answer is thus \(e^{\lim_{t \rightarrow 0} (t\ln(2t))} = e^0 = 1\). Therefore, the limit of the given function is: $$\lim_{(x, y) \rightarrow (0, 2)} (2xy)^{xy} = 1$$.