Question

Derivatives from limits The following limits represent \(f^{\prime}(a)\) for some function \(f\) and some real number \(a\) a. Find a possible function \(f\) and number \(a\). b. Evaluate the limit by computing \(f^{\prime}(a)\). $$\lim _{x \rightarrow 0} \frac{x+e^{x}-1}{x}$$

Short answer

The value of the limit is 1.

Step-by-step solution

Identify the limit as the derivative

The given limit can be recognized as the definition of a derivative of a function at a point, which is: $$f^{\prime}(a) = \lim _{x \rightarrow a} \frac{f(x)-f(a)}{x-a}$$ For our exercise, the given limit is: $$\lim _{x \rightarrow 0} \frac{x+e^{x}-1}{x}$$ Comparing the given limit with the definition, we can write it as: $$f^{\prime}(a) = \lim _{x \rightarrow 0} \frac{f(x)-f(0)}{x}$$ From this, we infer that \(a = 0\).

Identify a possible function \(f(x)\)

To find the function \(f(x)\), we need to find an antiderivative of the function in the limit. Observe that the numerator in the limit is the derivative of an unknown function: $$f(x) - f(a) = x + e^x - 1$$ So, we can find a possible \(f(x)\) by integrating the expression with respect to \(x\): $$f(x) = \int (x + e^x - 1) dx = \frac{x^2}{2} + e^x - x + C$$ Here, \(C\) is the constant of integration and can be determined by the condition \(f(a) = f(0)\), which we will find in the next step.

Find \(f(a)\) and \(C\)

We know that \(a = 0\), so we have: $$ f(0) = \frac{0^2}{2} + e^0 - 0 + C = C + 1$$ Now, we need to find \(f(0)\) so that we can determine the constant of integration \(C\). From our limit expression: $$f^{\prime}(a) = \lim _{x \rightarrow 0} \frac{x+e^{x}-1}{x} = \lim _{x \rightarrow 0} \frac{f(x)-f(0)}{x}$$ As \(x \rightarrow 0\), we can see that the limit is approaching: $$f^{\prime}(a) =\lim _{x \rightarrow 0} \frac{f(x) - f(0)}{x} = 1$$ Therefore, we can equate the limit as \(f(a)=1\), so we have: $$C + 1 = 1 \Rightarrow C = 0$$ Now, we have our function \(f(x)\): $$f(x) = \frac{x^2}{2} + e^x - x$$

Evaluate the limit

Now that we've found the function \(f(x)\) and our value of \(a = 0\), we can compute the limit to find the derivative, \(f^{\prime}(a)\): $$ f^{\prime}(a) = \lim _{x \rightarrow 0} \frac{x+e^{x}-1}{x}=1$$ So, we have found the function \(f(x) = \frac{x^2}{2} + e^x - x\) and the value of \(a = 0\). The limit, which represents the derivative \(f^{\prime}(a)\), is 1.

Key concepts

Limit

When we talk about limits in mathematics, we are exploring the behavior of a function as its input approaches a certain value. It essentially allows us to find the value that a function approaches as the input value gets closer and closer to some number. In this context, we are examining the limit:\[ \lim _{x \rightarrow 0} \frac{x+e^{x}-1}{x} \]
In this example, the limit helps us to determine the derivative of a function at a specific point. It's the foundation for understanding instantaneous rates of change, which are fundamental in calculus.
When the limit exists and is finite, it gives us a precise value or helps us determine the nature of the function at that point. In this exercise, the limit evaluates to 1, which serves to reveal the derivative at the point \(a = 0\). By properly understanding limits, we begin to unlock deeper concepts in calculus like continuity and derivatives.

Definition of Derivative

The concept of a derivative can be intimidating, but at its core, it's a measure of how a function changes as its input changes. The derivative of a function at a point gives us the slope of the tangent line to the curve at that point. Mathematically, the definition of a derivative is:\[ f^{\prime}(a) = \lim _{x \rightarrow a} \frac{f(x)-f(a)}{x-a} \]
In our exercise, the limit we are evaluating takes a form similar to this definition:\[ \lim _{x \rightarrow 0} \frac{x+e^{x}-1}{x} \]Here, it represents the derivative of a specific function at the point \(a = 0\). Recognizing this form led us to identify the involved function and evaluate it according to the definition of a derivative.Key aspects of derivatives include:

• Rate of Change: Reflects how much a function's output changes in response to changes in input.
• Critical Points: Where derivatives help determine local maximum and minimum points.

Understanding the definition of a derivative is essential for analyzing and solving many real-world problems that involve rates of change.

Function Integration

Integration, in contrast to differentiation, is essentially the process of finding a function from its derivative. It essentially answers the question: *"What function has this derivative?"* The integration process involves finding an antiderivative or integral for a given function.
In our problem, we were tasked with finding a function \(f(x)\) such that its derivative would match the numerator of our limit:\[ x + e^x - 1 \]By integrating this expression, we determined:\[ f(x) = \int (x + e^x - 1) \, dx = \frac{x^2}{2} + e^x - x + C \]
Key points about integration include:

• Antiderivatives: Functions that can be integrated to yield a given derivative.
• Constants of Integration: "C" often represents an unknown constant we determine through boundary conditions.
• Applications: Used in calculating areas under curves, total accumulated quantities, and solving differential equations.

Mastering integration techniques is vital because it allows us to reverse-engineer a problem from rates or gradients back into original functions or total values.

Evaluation of Limit

Evaluating a limit involves calculating the value that a function approaches as the input approaches a specific value. For our exercise, once we identified the derivative, we could assess the limit:\[ \lim _{x \rightarrow 0} \frac{x+e^{x}-1}{x} = 1 \]This evaluation of the limit confirmed the derivative value at the point.
When working with limits, several techniques can be highly useful:

• Simplification: Always simplify the expression, if possible, to make the limit easier to solve.
• L'Hôpital's Rule: Used if direct substitution results in an indeterminate form, like \(\frac{0}{0}\).
• Substitution: Sometimes substituting the direct limit value simplifies the expression.

Understanding how to evaluate limits not only helps with derivatives but also in determining the continuity of functions and solving complex calculus problems. Knowing these evaluation techniques can significantly simplify problem-solving in calculus scenarios.