Problem 47 - Precalculus with Limits

Question

In Exercises 47 and 48, find $$\textrm{(a)}\ \lim_{x \to 2}\ f(x), \quad \textrm{(b)} \lim_{x \to 2}\ g(x), \quad \textrm{(c)} \lim_{x \to 2}\ [f(x)g(x)], \quad \textrm{and (d)} \lim_{x \to 2}\ [g(x)-f(x)].$$ $$f(x)=x^3, \quad \quad g(x)=\dfrac{\sqrt{x^2 +5}}{2x^2}$$

Step-by-Step Solution

Answer

The limits are (a) 8, (b) 3/8, (c) 3, (d) -61/8.

Step 1: Finding the Limit of \( f(x) \)

Substitute \( x=2 \) into \( f(x)=x^3 \) to find that \( \lim_{x \to 2} f(x) = 2^3 = 8 \)

Step 2: Finding the Limit of \( g(x) \)

Substitute \( x=2 \) into \( g(x)=\frac{\sqrt{x^2 +5}}{2x^2} \) to find that \( \lim_{x \to 2} g(x) = \frac{\sqrt{2^2 +5}}{2*(2^2)} = \frac{\sqrt{9}}{8} = \frac{3}{8} \)

Step 3: Finding the Limit of the Product of \( f(x) \) and \( g(x) \)

Use the product of the individual limits to find the limit of the product, \( \lim_{x \to 2} [f(x)g(x)] = \lim_{x \to 2} f(x) * \lim_{x \to 2} g(x) = 8 * \frac{3}{8} = 3 \)

Step 4: Finding the Limit of the Difference of \( g(x) \) and \( f(x) \)

Use the difference of the individual limits to find the limit of the difference, \( \lim_{x \to 2} [g(x)-f(x)] = \lim_{x \to 2} g(x) - \lim_{x \to 2} f(x) = \frac{3}{8} - 8 = -\frac{61}{8} \)

Key Concepts

Limit of a Function

The limit of a function is a fundamental aspect in calculus. It helps us understand the behavior of a function as the input approaches a particular value.
When we state \( \lim_{x \to 2} f(x) \), it means we're interested in what value \( f(x) \) gets close to as \( x \) nears 2.
For the function \( f(x) = x^3 \), the limit as \( x \to 2 \) asks us to find \( f(2) \). This means replacing every \( x \) with 2 in \( x^3 \), resulting in \( 2^3 = 8 \).
Therefore, the function's limit at \( x = 2 \) is 8. This direct substitution works perfectly because \( x^3 \) is continuous and well-behaved at this point.\Whereas for \( g(x) = \frac{\sqrt{x^2 +5}}{2x^2} \), the same method applies.
By substituting \( x = 2 \) into \( g(x) \), we calculate \( \lim_{x \to 2} g(x) = \frac{\sqrt{2^2 +5}}{2*2^2} = \frac{\sqrt{9}}{8} = \frac{3}{8} \).
Limits assist us in predicting and estimating function values when calculating directly might be cumbersome or impossible.

Product of Limits

The property of the product of limits allows us to find the limit of a product of functions by separately finding the limits of the functions and multiplying them. This is much like the distributive property in arithmetic.
If \( f(x) \) and \( g(x) \) both have limits as \( x \to 2 \), we can compute their product's limit by using \( \lim_{x \to 2} [f(x)g(x)] = \lim_{x \to 2} f(x) \cdot \lim_{x \to 2} g(x) \).
Returning to our example: having calculated \( \lim_{x \to 2} f(x) = 8 \) and \( \lim_{x \to 2} g(x) = \frac{3}{8} \), we substitute these results into the property:

• \( \lim_{x \to 2} [f(x)g(x)] = 8 \cdot \frac{3}{8} = 3 \)

By breaking the problem into smaller piece by piece tasks, we simplify the process and ensure accuracy.

Difference of Limits

The difference of limits is another useful property, somewhat similar to the product property, but involving subtraction.
It states that if you have two functions, \( f(x) \) and \( g(x) \), with known limits, then the limit of their difference can be found by subtracting their individual limits.
Mathematically, it is expressed as \( \lim_{x \to 2} [g(x) - f(x)] = \lim_{x \to 2} g(x) - \lim_{x \to 2} f(x) \).
In our example, after determining \( \lim_{x \to 2} f(x) = 8 \) and \( \lim_{x \to 2} g(x) = \frac{3}{8} \), we apply the formula:

• \( \lim_{x \to 2} [g(x) - f(x)] = \frac{3}{8} - 8 = -\frac{61}{8} \)

This method is particularly beneficial when performing complex calculations by simplifying the process into manageable steps, ensuring a clearer path to the solution.