Question

Evaluate the following limits, where \(c\) and \(k\) are constants. \(\lim _{h \rightarrow 0} \frac{100}{(10 h-1)^{11}+2}\)

Short answer

Question: Evaluate the following limit: \(\lim_{h \rightarrow 0} \frac{100}{(10h-1)^{11}+2}\) Answer: The limit is equal to 100.

Step-by-step solution

Identify the type of limit

In this case, we have a simple limit because as \(h\) approaches 0, there is no indeterminate form. Hence, to evaluate the limit, we can simply substitute \(h=0\) in the expression.

Simplify the expression

Substitute \(h=0\) in the expression: \(\frac{100}{(10(0)-1)^{11}+2} = \frac{100}{(-1)^{11}+2}\)

Apply limit rules to evaluate the limit

As the expression is simplified, we can apply the limit rule and find the value: \(\lim_{h \rightarrow 0} \frac{100}{(10h-1)^{11}+2} = \frac{100}{(-1)^{11}+2} = \frac{100}{-1+2} = 100\) Therefore, the limit is: \(\lim_{h \rightarrow 0} \frac{100}{(10h-1)^{11}+2} = 100\)

Key concepts

Limit Evaluation

Understanding how to evaluate a limit is a fundamental aspect of calculus. When given a function, the goal is to determine what value the function approaches as the input approaches a specific point. In the presented problem, we are asked to evaluate \[ \lim_{h \rightarrow 0} \frac{100}{(10h-1)^{11}+2} \]The first step in limit evaluation is identifying the nature of the limit: is it a simple limit or does it form an indeterminate expression? If it's straightforward, as is this case, our task is simplified.

• For functions like this, evaluate by direct substitution whenever possible.
• This direct substitution approach is valid when the function doesn't produce undefined expressions like division by zero or any indeterminate form.

An important rule here is the simplification step, as seen with substituting \( h = 0 \), leading to:\[ \frac{100}{(-1)^{11}+2} = \frac{100}{-1 + 2} = 100 \] Thus, evaluating limits by substitution can make the process straightforward and effective.

Indeterminate Forms

Indeterminate forms are expressions encountered in calculus that don't initially provide enough information to determine a limit. Typical forms such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) often require special techniques, like L'Hôpital's Rule or algebraic manipulation, to evaluate properly.
In our exercise, the expression \( \lim_{h \rightarrow 0} \frac{100}{(10h-1)^{11}+2} \) is straightforward without any indeterminate form.

• Notice that as \( h \to 0 \), the denominator becomes \( (-1)^{11} + 2 \), which evaluates to a simple integer: \( -1 + 2 = 1 \).
• This indicates that the limit does not involve any unknown forms like \(\frac{0}{0}\). Thus, no additional manipulation is needed.

By recognizing these forms, we can efficiently determine whether advanced limit techniques are necessary, or if a simple substitution will suffice.

Substitution Method

The substitution method is a primary technique for determining limits when the function is continuous or doesn't involve indeterminate forms. It involves directly replacing the variable in the function with the value it is approaching as seen in our exercise.
This exercise demonstrates the substitution method well. Direct substitution becomes a powerful tool when the expression is simplified and results in a defined value.

• By substituting \( h = 0 \), we transformed the problem to evaluating: \( \frac{100}{(-1)^{11}+2} = \frac{100}{1} = 100 \).
• Substitution is perhaps the simplest method when applicable, streamlining the process of finding limits to basic arithmetic.

Unlike using L'Hôpital's Rule or algebraic simplification for indeterminate forms, substitution is quick and effective when applicable. Properly identifying when you can employ this approach can save both time and effort.