Question

Continuous Functions In Exercises 73 and \(74,\) find the value of \(c\) that makes the function continuous at \(x=0\). \(f(x)=\left\\{\begin{array}{ll}\left(e^{x}+x\right)^{1 / x}, & x \neq 0 \\ c, & x=0\end{array}\right.\)

Short answer

The value of \( c \) that makes the function continuous at \( x = 0 \) is \( e^2 \)

Step-by-step solution

Identification of indeterminate form

First we would like to know the limit of \( (e^x + x) ^ {1 / x} \) when \( x \) approaches 0. We try to substitute \( x = 0 \) directly, both in numerator and denominator. We get \( (e^0 + 0) ^ {1 / 0} \), which simplifies to 1 raised to power infinity. This is an indeterminate form, so we cannot directly evaluate the limit.

Transform the indeterminate form to \( 0 / 0 \) form

We can rewrite the limit in logarithmic form: \( \lim_{x \to 0} ln \left( (e^{x}+x) ^ {1 / x} \right) \) which further simplifies to \( \lim_{x \to 0} \frac{ln(e^{x}+x)}{x} \). Now we get \( 0/0 \) form while substituting \( x=0 \), which is a prerequisite to apply l'Hopital's rule.

Apply l'Hopital's Rule and evaluate the limit

Applying l'Hopital’s rule, we get \( \lim_{x \to 0} \frac{e^{x} + 1}{1} \). Now substituting \( x=0 \), we obtain the limit value is \( 2 \). Thus, \( \lim_{x \to 0} (e^{x} + x) ^ {1 / x} \) is \( e^2 \).

Decide the value of \( c \)

For the function to be continuous at \( x = 0 \), the function value at \( x = 0 \) should be the same as the limit when \( x \) approaches 0. We found out that the limit is \( e^2 \). Thus, \( c \) should be \( e^2 \) for the function to be continuous at \( x = 0 \).

Key concepts

Indeterminate Forms

When evaluating limits in calculus, we sometimes encounter expressions that don't produce a clear numeric value. These are known as indeterminate forms. One common example is '1 raised to the power of infinity' (\(1^{\text{infinity}}\)), which is deceptive in appearance because it seems definable but upon closer analysis, possesses no clear value.

The challenge in calculus is that these forms can actually represent a range of different values, depending on the context. To make sense of these, we have to manipulate the expressions into a form amenable to further analysis and use specialized methods, such as l'Hopital's Rule, to find the limit.

L'Hopital's Rule

L'Hopital's Rule is a hugely valuable tool in calculus for dealing with limits that produce indeterminate forms like \(0/0\) or \(\infty/\infty\). When direct substitution in a limit leads to an indeterminate form, we can apply this rule, which states that under certain conditions, the limit of the ratio of two functions can be found by taking the limit of the ratio of their derivatives.

To apply L'Hopital's Rule, we differentiate the numerator and the denominator separately, then take the limit of the new ratio. This method simplifies the finding of limits and, as shown in the provided exercise, changes an indeterminate form into a determinate one, allowing us to evaluate the limit and hence the continuity of a given function.

Limits

In calculus, a limit describes the value that a function approaches as the input (or the 'x' value) approaches a given value. Limits are foundational to calculus as they form the basis for defining derivatives and integrals.

In the context of our exercise, we were interested in the limit of \( (e^x + x) ^ {1 / x} \) as \( x \) approaches 0. Limits help us understand the behavior of functions at points that might not be easily evaluated by direct substitution, such as points of discontinuity or, as our exercise illustrated, points involving indeterminate forms.

Calculus

Calculus is a branch of mathematics that studies how things change. It provides a framework for modeling systems in which there is change, and a way to deduce the predictions of such models. There are two major divisions in calculus: differential calculus concerning rates of change and slopes of curves, and integral calculus concerning accumulation of quantities and the areas under and between curves.

The problems in calculus often involve finding the limit of a function to understand its behavior and continuity. As we saw in our exercise, a concrete grasp of calculus concepts such as limits and the toolset provided by rules like L'Hopital's are essential for solving a wide range of mathematical problems.