Problem 53 - Calculus: Early Transcendentals

Question

Find the following limits or state that they do not exist. Assume \(a, b, c,\) and k are fixed real numbers. $$\lim _{h \rightarrow 0} \frac{(5+h)^{2}-25}{h}$$

Step-by-Step Solution

Answer

Answer: The limit of the function as h approaches 0 is 10.

Step 1: Simplify the given expression

We have the expression: \(\frac{(5+h)^{2}-25}{h}\). Start by expanding the numerator: \[((5+h)^2 - 25 = (5^2 + 2(5)(h) + h^2) - 25\] This simplifies to: \[(25 + 10h + h^2) - 25\] Now, we can cancel out the 25's, which results in the expression: \[10h + h^2\]

Step 2: Rewrite the expression as a product of two factors

Now we rewrite the simplified expression as a product of two factors with a factor h in each term: \[h(10 + h)\]

Step 3: Factor out h from the numerator and simplify the expression

Now divide both the numerator and the denominator by h: \[\frac{h(10 + h)}{h}\] Cancelling out h's, we get: \[10+h\]

Step 4: Evaluate the limit as h approaches 0

Now we can easily evaluate the limit as h approaches 0 by substituting 0 into the expression: \[\lim_{h \rightarrow 0} (10 + h) = 10+0\] So the limit is: \[\lim_{h \rightarrow 0} \frac{(5+h)^{2}-25}{h} = 10\]

Key Concepts

Limit of a Function

The concept of a 'limit of a function' is foundational in calculus and helps us understand the behavior of functions as they approach a specific point. For students tackling limits for the first time, it can be quite challenging. Think of a limit as the value that a function 'approaches' as the input (or the independent variable) gets closer to some number. Limits are critical for defining derivatives, integrals, and the continuity of functions.

For example, in the given exercise \[\lim _{h \rightarrow 0} \frac{(5+h)^{2}-25}{h}\], we are asked to find the limit of a rational function as \(h\), which plays the role of the independent variable here, approaches zero. This approach is fundamental in the definition of the derivative, which in essence measures the rate of change of a function.

Factoring Algebraic Expressions

Factoring is a powerful tool in algebra that simplifies expressions and makes problems more manageable, particularly in calculus when dealing with limits and derivatives. The process involves writing an expression as a product of its factors. Factoring is particularly helpful when evaluating limits of algebraic expressions where direct substitution is not possible due to forms like \(\frac{0}{0}\).

In the context of our problem, we see this process in action as the square of a binomial is expanded and then simplified. Once the numerator is expanded from \((5+h)^{2}\), we get \(25 + 10h + h^2 - 25\), and subsequently, we factor out an \(h\), to rewrite the expression as \(h(10 + h)\). At this stage, the apparent discontinuity when \(h = 0\) is resolved by canceling out the common factors in the numerator and denominator, which is a common strategy in evaluating limits.

Evaluating Limits

The final step in finding the limit of an algebraic expression involves directly evaluating the simplified expression by plugging in the value towards which the variable approaches. This method is straightforward if the function is continuous. However, if a direct substitution leads to an indeterminate form, such as \(\frac{0}{0}\), other strategies such as factoring, using conjugates, or applying L'Hôpital's rule might be necessary.

In the given exercise, after factoring and simplifying the expression to \(10 + h\), we easily evaluated the limit by substituting \(h = 0\) to get 10. This outcome tells us that as \(h\) gets infinitesimally close to zero, the expression \(\frac{(5+h)^{2}-25}{h}\) approaches 10. Thus, understanding how to evaluate limits is not only about becoming comfortable with algebraic manipulation but also about grasping the 'big picture' of what it signifies about the behavior of functions.