Question

Calculating limits exactly Use the definition of the derivative to evaluate the following limits. $$\lim _{h \rightarrow 0} \frac{(3+h)^{3+h}-27}{h}$$

Short answer

Answer: The limit of the function is 0.

Step-by-step solution

Rewrite the function using exponent properties

The given function is: $$\lim _{h \rightarrow 0} \frac{(3+h)^{3+h}-27}{h}$$ The expression \((3 + h)^{3 + h}\) can be separated into two parts using the exponent properties: \((3 + h)^3 \cdot (3 + h)^h\). So, the function becomes: $$\lim _{h \rightarrow 0} \frac{(3+h)^3(3+h)^h - 27}{h}$$

Factor out common terms

We can factor out \((3 + h)^3\) from the numerator: $$\lim _{h \rightarrow 0} \frac{(3+h)^3[(3+h)^h - 1]}{h}$$

Apply L'Hôpital's Rule

As this is a limit in the 0/0 indeterminate form, we can apply L'Hôpital's Rule. To apply the rule, we need to differentiate both the numerator and the denominator regarding h: $$\lim _{h \rightarrow 0} \frac{d}{dh}[(3+h)^3[(3+h)^h - 1]] \bigg/ \frac{d}{dh}[h]$$ Differentiate the numerator using the product and chain rules: $$\lim _{h \rightarrow 0} \frac{3(3+h)^2[(3+h)^h - 1] + (3+h)^3\cdot h\cdot(3+h)^{h-1}\cdot\ln(3+h)}{1}$$

Evaluate the limit

Now, we can substitute h with 0 and evaluate the limit: $$\frac{3(3)^2[(3)^0 - 1] + (3)^3\cdot 0\cdot(3)^{-1}\cdot\ln(3)}{1}$$ $$\frac{3\cdot 9[1 - 1]}{1}$$ So, the limit is equal to 0: $$\lim _{h \rightarrow 0} \frac{(3+h)^{3+h}-27}{h} = 0$$

Key concepts

limits

A limit is an essential concept in calculus that describes the behavior of a function as its input approaches some value. It helps us understand what value a function is heading towards even if it doesn't exactly reach it. In this exercise, we evaluate the limit as \( h \) approaches zero.

• The interpretation involves finding the value that the expression \( \frac{(3+h)^{3+h}-27}{h} \) tends to as \( h \) becomes very small.
• Limits are vital for calculating derivatives, which signify rate of change, providing the foundation for much of calculus.
• In expressions with indeterminate forms, such as \( \frac{0}{0} \), limits allow us to apply analytical techniques like L'Hôpital's Rule to find actual values.

Understanding limits is critical for tackling more complicated calculus problems and plays a huge role in finding the derivatives of functions.
This particular limit guides us through simplifying and differentiating the function using L'Hôpital's Rule. This step is crucial where a direct substitution would not work due to the form of the function.

L'Hôpital's Rule

L'Hôpital's Rule is incredibly useful for solving limits that result in indeterminate forms such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). This rule allows us to take the derivative of the function's numerator and denominator separately to find the limit.

• It simplifies complex limit expressions, like those seen in calculus involving exponential and polynomial functions.
• In our exercise, the expression \( \frac{(3+h)^{3+h} - 27}{h} \) forms a \( \frac{0}{0} \) when \( h = 0 \), making it a prime candidate for this rule.
• By differentiating each part, we transform an indeterminate form into something that can be resolved through straightforward evaluation.

Applying L'Hôpital's Rule provides an analytical rather than intuitive path to finding limits, making it a powerful tool in calculus.
Through this rule, complex problems involving limits become sufficiently simplified, such as in this exercise, enabling the derivation of a definite solution.

chain rule

The Chain Rule is a fundamental principle in calculus used for differentiating compositions of functions. It's instrumental when dealing with functions like \( (3+h)^h \) in our exercise.

• The rule states that to differentiate a composite function, you take the derivative of the outer function and multiply it by the derivative of the inner function.
• This multi-step process allows for smooth differentiation of nested functions that hinder direct application of basic differentiation rules.
• In our instance, differentiating \( (3+h)^h \) reveals its role, highlighting how changes in \( h \) influence complete expressions.

Comprehending the Chain Rule is crucial for managing various complex derivatives and significantly impacts calculus problem-solving.
Utilizing the Chain Rule is a key component of this solution, aiding in the breakdown of one of the core parts of the function before applying L'Hôpital's Rule.

product rule

The Product Rule in calculus is vital when you're trying to differentiate products of two or more functions. This rule is specially applied when you encounter terms like \( (3+h)^3 [(3+h)^h - 1] \) in our given expression.

• It requires taking the derivative of the first function and multiplying it by the second function as is, then adding the product of the original first function and the derivative of the second function.
• This rule lets us perform operations on expressions where overlapping functions multiply each other, like seen here.
• By applying the rule, each component's derivative is considered, ensuring an accurate representation of the overall change in function value.

Understanding the Product Rule is essential for proper differentiation in problems involving multiple functions interacting multiplicatively.
In the context of our exercise, it helps to dissect the complexity of the original limit expression when used with L'Hôpital's Rule, clarifying the derivative processes needed for each part.