Problem 26 - Calculus Early Transcendentals: Pearson New International Edition

Question

, find the limits if \(\lim _{x \rightarrow a} f(x)=3\) and \(\lim _{x \rightarrow a} g(x)=-1(\) see Example 4) \(.\) $$ \lim _{x \rightarrow a} \frac{2 f(x)-3 g(x)}{f(x)+g(x)} $$

Step-by-Step Solution

Answer

The limit is \(\frac{9}{2}\).

Step 1: Apply Limit Properties

We start by applying the properties of limits to separate the limits of the numerator and denominator. According to limit properties: \[\lim_{x \to a} \frac{2f(x) - 3g(x)}{f(x) + g(x)} = \frac{\lim_{x \to a} (2f(x) - 3g(x))}{\lim_{x \to a} (f(x) + g(x))}\]

Step 2: Calculate the Limit of the Numerator

Calculate the limit of the numerator: \[\lim_{x \to a} (2f(x) - 3g(x)) = 2 \cdot \lim_{x \to a} f(x) - 3 \cdot \lim_{x \to a} g(x)\]Substitute the given limits:\[2 \times 3 - 3 \times (-1) = 6 + 3 = 9\]

Step 3: Calculate the Limit of the Denominator

Calculate the limit of the denominator: \[\lim_{x \to a} (f(x) + g(x)) = \lim_{x \to a} f(x) + \lim_{x \to a} g(x)\]Substitute the given limits:\[3 + (-1) = 2\]

Step 4: Divide the Limits

Now, substitute the values obtained in Steps 2 and 3 back into the equation from Step 1:\[\frac{\lim_{x \to a} (2f(x) - 3g(x))}{\lim_{x \to a} (f(x) + g(x))} = \frac{9}{2}\]

Step 5: Conclusion

The limit of the given expression as \(x \to a\) is \(\frac{9}{2}\).

Key Concepts

Understanding Properties of Limits

In calculus, the properties of limits help us evaluate complex limit expressions by breaking them down into simpler parts. These properties provide the foundation for separating limits into more manageable computations.
For instance, if the limit of a combination of functions can be determined, this can often be done by evaluating each function's limit separately and then combining them back together. This is essential in understanding how to manipulate expressions, especially when both polynomials and rational functions are involved.
Key properties include:

• Limit of a Sum: The limit of a sum is the sum of the limits.
• Limit of a Difference: The limit of a difference is the difference of the limits.
• Limit of a Product: The limit of a product is the product of the limits.
• Limit of a Quotient: The limit of a quotient is the quotient of the limits, provided the limit of the denominator isn't zero.

These properties simplify the calculation of limits by allowing you to evaluate parts of the expression individually. This is precisely the method applied to our exercise, where the entire expression is broken down into the limits of its numerator and denominator.

Numerator and Denominator Limits

In evaluating fractional limits, it's crucial to isolate the limit of both the numerator and the denominator. The numerator in our exercise is the expression \(2f(x) - 3g(x)\), and the denominator is \(f(x) + g(x)\).
By applying the properties of limits, we split these into more manageable parts. Remember, the goal is to replace each function with its limit at a particular point, making it possible to simplify the fraction altogether.

• For the numerator, you calculate: \(\lim_{x \to a} \left( 2f(x) - 3g(x) \right) = 2 \cdot \lim_{x \to a} f(x) - 3 \cdot \lim_{x \to a} g(x)\)
• For the denominator, it's: \(\lim_{x \to a} \left( f(x) + g(x) \right) = \lim_{x \to a} f(x) + \lim_{x \to a} g(x)\)

After substituting the given limits, this helps you find concrete values for both the numerator and denominator. This allows a clearer path to solving for the overall limit.

Substituting Limits

After evaluating individual limits for the numerator and denominator, the next step is substituting these limit values back into the partial expressions. Substitution transforms the limit from abstract forms into precise numbers that are easier to understand.
In our example, once you have:

• The limit of the numerator as \(9\)
• The limit of the denominator as \(2\)

it's straightforward to substitute these values back into the fraction. This substitution step is a pivotal moment where the theoretical work transitions into a concrete solution. Pay close attention to signs and operations during this step to keep calculations accurate and meaningful. Ultimately, this direct substitution leads to the simplified form of the solution.

Evaluating Fractional Limits

When faced with fractional limits, such as in our example, it's important to ensure that the denominator's limit isn't zero, as dividing by zero is undefined. In our exercise, after we've calculated, the denominator limit is \(2\), a non-zero value, allowing us to correctly finish the limit expression.
Learning how to handle fractional limits involves:

• Careful computation of numerator and denominator limits separately
• Substitution of known limit values back into the fraction
• Ensuring no division by zero occurs, which could disrupt finding a valid limit

Once all these steps are successfully navigated, the result is a definitive answer for the limit, in this case, \(\frac{9}{2}\). Understanding this process helps demystify the steps involved in effectively working with fractions in calculus, reinforcing both your analytical skills and confidence in solving similar problems.