Question

Show that the indeterminate forms \(0^{\circ}\) \(\infty^{0},\) and \(1^{\infty}\) do not always have a value of 1 by evaluating each limit. (a) \(\lim _{x \rightarrow 0^{+}} x^{\ln 2 /(1+\ln x)}\) (b) \(\lim _{x \rightarrow \infty} x^{\ln 2 /(1+\ln x)}\) (c) \(\lim _{x \rightarrow 0}(x+1)^{(\ln 2) / x}\)

Short answer

The limits for the forms \(0^0\), \(\infty^0\), and \(1^\infty\) respectively are 1, 1, and 2.

Step-by-step solution

Evaluate Limit (a)

The limit is in the form of \(0^0\), which is an indeterminate form. To approach this, apply logarithms. Firstly, re-write the limit in exponential form as follows: \(\lim _{x \rightarrow 0^{+}} e^{\ln x \cdot \frac{\ln 2}{1+\ln x}}\). Now, apply the logarithmic rule \(x*y = x+y\). After applying, you get \(\lim _{x \rightarrow 0^{+}} e^{\frac{\ln 2 \cdot \ln x}{1+\ln x}}\). Next, apply l'Hopital's rule since the limit is in the form \(\frac{0}{0}\). After evaluating the limit, you get \(e^{0}\), which equals 1.

Evaluate Limit (b)

The limit is in the form of \(\infty^0\), which is also an indeterminate form. Apply logarithms just like in step 1 and rewrite the limit to its exponential form as \(\lim _{x \rightarrow \infty} e^{\ln x \cdot \frac{\ln 2}{1+\ln x}}\). Now, use \(\ln(u*v) = \ln(u) + \ln(v)\) to get \(\lim _{x \rightarrow \infty} e^{\frac{\ln 2 \cdot \ln x}{1+\ln x}}\). Again, apply l'Hopital's rule as the limit is now in form \(\frac{\infty}{\infty}\). After evaluating, the limit yields \(e^{0}\), which equals 1.

Evaluate Limit (c)

The limit is in the form of \(1^\infty\), which is an indeterminate form. Apply logarithms and rewrite the limit to exponential form as \(\lim _{x \rightarrow 0} e^{\ln (x+1) \cdot \frac{\ln 2}{x}}\). Now, use \(\ln(u*v) = \ln(u) + \ln(v)\) and you get \(\lim _{x \rightarrow 0} e^{\frac{\ln 2 \cdot \ln (x+1)}{x}}\). Apply l'Hopital's rule, and after evaluating, the limit turns out to be \(e^{\ln 2}\), which equals 2.

Key concepts

Limits in Calculus

Understanding limits is crucial when it comes to studying calculus. At its core, a limit describes the value that a function approaches as the input (or argument of the function) approaches some value. Limits are foundational for defining derivatives and integrals, which form the pillars of calculus.

When dealing with limits, we sometimes face what are called 'indeterminate forms'. These are expressions we can't directly infer the behavior of without doing some additional work. Classic examples include \(0^0\text{, } 0 \cdot \infty\text{, and } \frac{\infty}{\infty}\). In the exercise given, we encounter limits that present indeterminate forms like \(0^0\text{, } \infty^0\text{, and } 1^\infty\), which don't have straightforward values and can't be simplistically said to be equal to 1.

For understanding such limits, we can't rely on our intuition or simple plug-and-play methods; we must use more sophisticated techniques like l'Hopital's rule (which we'll discuss next), or algebraic manipulation that might involve logarithms, as shown in the exercise solutions.

L'Hopital's Rule

When two functions, \(f(x)\) and \(g(x)\), approach 0 or infinity as \(x\) approaches a certain value and you're trying to find the limit of their quotient, \(\frac{f(x)}{g(x)}\), you might end up with an indeterminate form. In such cases, l'Hopital's Rule is a powerful tool. It states that if the limit of \(\frac{f(x)}{g(x)}\) as \(x\) approaches a value \(c\) results in \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then this limit is the same as the limit of the derivatives of these functions, \(\frac{f'(x)}{g'(x)}\), provided this latter limit exists.

In our exercise, we apply l'Hopital's rule to evaluate the complicated limits. By converting to exponential forms and using properties of logarithms, we established the limits as ratios that yield indeterminate forms. Then, by differentiating the numerator and denominator separately and finding their limits, we simplified complex expressions to find the precise values of the indeterminate forms.

Exponential Functions

Exponential functions, written generally as \(a^x\), are vital in both theoretical and applied mathematics. They display consistent growth or decay patterns and pop up across various disciplines, from finance to physics.

The number \(e\), approximately equal to 2.718, is a special base for exponential functions due to its unique properties in calculus, particularly regarding differentiation and integration. In the given exercise, we are dealing with the natural exponential function, \(e^x\), which is its own derivative and integral – a remarkable trait that simplifies many calculus problems.

In parts (a) and (b) of our exercise, after applying logarithms and l'Hopital's rule, we find the limits are \(e^0\) which simplifies to 1. This illustrates that \(0^0\) and \(\infty^0\) do not necessarily lead to the value of 1 inherently. In part (c), we see that the indeterminate form \(1^\infty\) results in \(e^{\ln 2}\), which simplifies to 2, further debunking the common misconception that such indeterminate forms result in a value of 1.