Question

Let $$f(x)=\left\\{\begin{array}{ll} \frac{e^{x}-1}{x}, & \text { if } x \neq 0 \\ c, & \text { if } x=0 \end{array}\right.$$ What value of \(c\) makes \(f(x)\) continuous at \(x=0\) ?

Short answer

The value of \( c \) that makes \( f(x) \) continuous at \( x = 0 \) is 1.

Step-by-step solution

Define Continuity at a Point

For a function to be continuous at a point, the limit of the function as it approaches the point from either side must equal the function's value at that point. Specifically, for continuity at \( x = 0 \), \( \lim_{x \to 0} f(x) = f(0) \). In this case, this means \( \lim_{x \to 0} f(x) = c \).

Express the Limit Using the Given Function

Since \( f(x) = \frac{e^x - 1}{x} \) when \( x eq 0 \), we need to find \( \lim_{x \to 0} \frac{e^x - 1}{x} \). This will determine the value of \( c \) that makes \( f(x) \) continuous at \( x = 0 \).

Apply L'Hôpital's Rule

The limit \( \lim_{x \to 0} \frac{e^x - 1}{x} \) is an indeterminate form of \( \frac{0}{0} \). To solve it, we can use L'Hôpital's Rule, which states that for indeterminate forms, \( \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \) provided the latter limit exists. Here, \( f(x) = e^x - 1 \) and \( g(x) = x \). The derivatives are \( f'(x) = e^x \) and \( g'(x) = 1 \).

Evaluate the Limit Using Derivatives

Now, apply L'Hôpital's Rule:\[ \lim_{x \to 0} \frac{f'(x)}{g'(x)} = \lim_{x \to 0} \frac{e^x}{1} = \lim_{x \to 0} e^x = e^0 = 1. \]Therefore, the limit is 1.

Determine the Value of c

For \( f(x) \) to be continuous at \( x = 0 \), we need \( \lim_{x \to 0} f(x) = f(0) = c \). From our calculation, \( \lim_{x \to 0} \frac{e^x - 1}{x} = 1 \). Therefore, \( c = 1 \).

Key concepts

Limits

Limits are a fundamental concept in calculus, representing the value that a function approaches as the input approaches a particular point. They help define and understand behavior at specific points, especially where a function might not be directly defined. In the problem of finding continuity of a function, limits are crucial. For instance, the exercise involves the function \[ f(x) = \frac{e^x - 1}{x} \] and seeks the value of \(c\) that ensures continuity at \(x = 0\). Here, you calculate \[ \lim_{x \to 0} \frac{e^x - 1}{x} \]to determine the function's behavior near zero. Limits are analyzed by observing function values as the input gets closer and closer to the point of interest (approaching either from the left or the right). In this case, we evaluate the limit approaching zero to ensure the function remains consistent there. This work reveals the value needed for the function to be seamlessly continuous at a specific point.

L'Hôpital's Rule

L'Hôpital's Rule is a handy tool in calculus for figuring out limits involving indeterminate forms such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).When you encounter a limit like \[ \lim_{x \to a} \frac{f(x)}{g(x)} \]and it results in an indeterminate form, L'Hôpital's Rule allows you to take the derivatives of \(f(x)\) and \(g(x)\) and then find the limit of their ratio instead:\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \]This strategy only works if the new limit exists.In the exercise, \[ \lim_{x \to 0} \frac{e^x - 1}{x} \]is indeterminate and needs L'Hôpital's Rule to solve. Finding the derivatives, \( f'(x) = e^x \) and \( g'(x) = 1 \), turns the problem into:\[ \lim_{x \to 0} \frac{e^x}{1} = \lim_{x \to 0} e^x = e^0 = 1 \]Thanks to L'Hôpital's Rule, solving tricky limits becomes more approachable by converting them into a simpler form.

Indeterminate Forms

Indeterminate forms often appear in calculus when evaluating limits that don't straightforwardly resolve.These forms might initially seem like they're leading to a contradiction, such as \( \frac{0}{0} \), \( 0 \times \infty \), or \( \infty - \infty \), where a direct substitution leaves ambiguity about the function's value.In the given exercise, the limit to check continuity results in \[ \lim_{x \to 0} \frac{e^x - 1}{x} = \frac{0}{0}, \]making it an indeterminate form. L'Hôpital's Rule, among other techniques, comes into play to resolve these problems effectively. By determining the derivatives of the numerator and denominator, and then taking the new limit, you can gain a clearer answer about the situation. Understanding indeterminate forms helps reinforce the ideas of flexibility and adaption in solving calculus problems, especially when the straightforward path is blocked.