Question

Calculating limits exactly Use the definition of the derivative to evaluate the following limits. $$\lim _{x \rightarrow 2} \frac{5^{x}-25}{x-2}$$

Short answer

Question: Evaluate the limit $$\lim_{x \rightarrow 2} \frac{5^{x}-25}{x-2}$$. Answer: The value of the limit is $$25 \log 5$$.

Step-by-step solution

Rewrite the expression using the definition of the derivative

Recall that the definition of the derivative is: $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ We are given the expression: $$\lim_{x \rightarrow 2} \frac{5^{x}-25}{x-2}$$ It looks like a derivative, except the variable is x, not h. We can rewrite this expression in terms of the definition of the derivative using substitution. Let's set: $$f(x) = 5^x$$ Then let's consider the point when x = 2, so our expression is: $$\lim_{x \rightarrow 2} \frac{f(x)- f(2)}{x-2}$$

Differentiate using the definition

Now we're set to differentiate with respect to x using the definition of the derivative: $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ So, we will first find \(f(x+h)\), then plug it into the definition and find the limit as h approaches 0. $$f(x+h) = 5^{x+h}$$ Now we can write the definition of the derivative for this function: $$f'(x) = \lim_{h\to 0} \frac{5^{x+h} - 5^x}{h}$$

Apply the limit at x = 2

We are tasked with finding the value of this limit when x = 2. So let's plug in x = 2 into our derivative expression and simplify: $$f'(2) = \lim_{h\to 0} \frac{5^{2+h} - 5^2}{h}$$ Now we are ready to find the limit of this expression as h approaches 0: $$f'(2) = \lim_{h\to 0} \frac{5^{2+h} - 25}{h}$$

Use limit properties to evaluate the limit

To evaluate this limit, we can use the exponential limit property: $$\lim_{h\to 0} a^{f(h)} = a^{\lim_{h\to 0} f(h)}$$ So in our case, we have: $$f'(2) = \lim_{h\to 0} \frac{5^{2+h} - 25}{h} = \frac{5^{\lim_{h\to 0} (2+h)} - 25}{\lim_{h\to 0} h}$$

Use L'Hospital's Rule

We still have a limit of the form \(\frac{0}{0}\). To resolve this, we can use L'Hospital's Rule, which states that for a limit of the form \(\frac{f(x)}{g(x)}\), where the limit of both f(x) and g(x) is 0, we have: $$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$$ In our case, we apply L'Hospital's rule on our expression by differentiating with respect to h both the numerator and denominator: $$f'(2) = \lim_{h \to 0} \frac{\frac{\partial}{\partial h} (5^{2+h} - 25)}{\frac{\partial}{\partial h} h}$$ The derivative with respect to h of the numerator is: $$\frac{\partial}{\partial h} (5^{2+h} - 25) = 5^2 \cdot \log 5 \cdot 5^h$$ The derivative with respect to h of the denominator is just 1, so we have: $$f'(2) = \lim_{h \to 0} 5^2 \cdot \log 5 \cdot 5^h$$

Evaluate the final limit

Now we evaluate the limit as h approaches 0: $$f'(2) = 5^2 \cdot \log 5 \cdot 5^0 = 25 \cdot \log 5 \cdot 1 = 25 \log 5$$ So, the exact value of the given limit is: $$\lim_{x \rightarrow 2} \frac{5^x - 25}{x-2} = 25 \log 5$$

Key concepts

Definition of the Derivative

Understanding the definition of the derivative is crucial in calculus, especially when handling complex limits. The derivative of a function at a certain point describes the rate at which the function's value changes at that point. Represented as \( f'(x) \), it is formally defined by the limit of the difference quotient as the increment approaches zero:
\[ f'(x) = \text{lim}_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h} \]
This definition is at the heart of calculus as it connects the concept of the limit to the slope of the tangent line to the curve of the function at any point.

Exponential Functions

Exponential functions are a category of mathematical functions of the form \( f(x) = a^x \), where the base \( a \) is a constant and the exponent \( x \) is a variable. These functions are characterized by a constant rate of growth or decay. In our exercise, the function presented is \( 5^x \), which is an exponential function with a base of 5.
When dealing with limits involving exponential functions, it's important to apply properties of exponents and sometimes, as in the solution, L'Hospital's rule is employed to resolve forms that are indeterminate, like \( \frac{0}{0} \).

L'Hospital's Rule

L'Hospital's Rule is an invaluable tool for determining the limit of a quotient of two functions that yields an indeterminate form such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). The rule states that the limit of the ratio of the functions can be found by taking the limit of the ratio of their derivatives, assuming that the limit exists:
\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \]
This approach simplifies the calculation of limits that would otherwise be challenging to evaluate directly. It's essential, however, to confirm that using the rule is appropriate for the functions at hand.

Evaluating Limits

Evaluating limits is a fundamental task in calculus. Determining the value that a function approaches as the input approaches a certain point allows for the understanding of function behavior near that point. Though some limits can be computed directly by substitution, in cases where the function is undefined or yields an indeterminate form at the point, advanced techniques such as factoring, rationalizing, or applying L'Hospital's Rule, as in the above exercise, are used to determine the limit.
Mastering the evaluation of limits is essential because it underpins other calculus concepts such as continuity, derivatives, and integrals.