Problem 69 - Calculus Volume 3

Question

For the following exercises, evaluate the limits at the indicated values of \(x\) and \(y\). If the limit does not exist, state this and explain why the limit does not exist. $$\lim _{(x, y) \rightarrow(2,5)}\left(\frac{1}{x}-\frac{5}{y}\right)$$

Step-by-Step Solution

Answer

The limit is \(-0.5\).

Step 1: Substitute the values into the expression

To evaluate the limit, substitute the given values of \(x\) and \(y\) into the expression. The expression is \(\frac{1}{x} - \frac{5}{y}\). Therefore, we need to plug \(x = 2\) and \(y = 5\) into this expression.

Step 2: Calculate \(\frac{1}{x}\) with \(x = 2\)

Substitute \(x = 2\) into the first term of the expression, \(\frac{1}{x}\). This gives \(\frac{1}{2}\).

Step 3: Calculate \(\frac{5}{y}\) with \(y = 5\)

Now substitute \(y = 5\) into the second term of the expression, \(\frac{5}{y}\). This results in \(\frac{5}{5}\), which equals 1.

Step 4: Subtract the two results

Now, subtract the result from Step 3 from the result in Step 2. This means calculating \(\frac{1}{2} - 1\).

Step 5: Simplify the result

Simplify \(\frac{1}{2} - 1\) to obtain the final value. \(\frac{1}{2}\) is the same as \(0.5\), and \(1\) is the same as \(1.0\), so subtracting gives \(-0.5\).

Key Concepts

Evaluating Limits

Evaluating limits is a fundamental concept in calculus. It involves finding the value that a function approaches as the input (or variable) approaches a specific point. Understanding limits is essential because it helps in analyzing the behavior of functions at boundaries or points of interest.
To evaluate a limit, you often substitute values into the function. If direct substitution leads to a valid numerical answer, that's the limit. Sometimes, substitution might not work if it leads to undefined expressions, like division by zero.
In the exercise, the limit evaluates at the point \((x, y) = (2, 5)\). This means we're interested in what the expression \(\frac{1}{x} - \frac{5}{y}\) approaches as \(x\) gets closer to 2 and \(y\) gets closer to 5. By substituting these values directly, we avoided complex calculations and easily found the limit.

Substitution Method

The substitution method is often the first technique we use for evaluating limits. Here's how it works:

• Take the given values and plug them directly into the function.
• If the function provides a defined outcome (not undefined or infinite), that result is the limit.

This method is especially useful for functions that are continuous, meaning there aren't any breaks or holes at the point you're evaluating.
In our specific exercise, we used the substitution method to evaluate the limit \(\lim _{(x, y) \rightarrow(2,5)}\left(\frac{1}{x}-\frac{5}{y}\right)\).

• We substituted \(x = 2\) into \(\frac{1}{x}\) and got \(\frac{1}{2}\).
• We then substituted \(y = 5\) into \(\frac{5}{y}\) and got \(1\).
• Finally, we subtracted these results to find the limit.

The substitution method made it straightforward to find the limit as \(-0.5\).

Calculus Exercises

Calculus exercises, like evaluating limits, are designed to reinforce your understanding of mathematical concepts. They often involve applying specific methods or formulas to solve problems.
These exercises begin by presenting a function and a point where you need to evaluate the limit. You then use appropriate calculus techniques, such as the substitution method, to find an answer. Calculus exercises help in:

• Enhancing problem-solving skills.
• Building familiarity with applying calculus techniques to real problems.
• Developing a systematic approach to mathematical reasoning and analysis.

The original exercise, asking to evaluate \(\lim _{(x, y) \rightarrow(2,5)}\left(\frac{1}{x}-\frac{5}{y}\right)\), is typical of calculus practice. Such exercises help ensure students gain confidence in tackling similar problems independently.