Question

Evaluate the following limits. $$\lim _{x \rightarrow 0^{+}} x^{1 /(1+\ln x)}$$

Short answer

Answer: The limit of the expression \(x^{1 /(1+\ln x)}\) as x approaches 0 from the positive side is \(e\).

Step-by-step solution

Rewrite the expression

To rewrite the expression to be more suitable for applying L'Hôpital's rule, we can use the property of natural logarithms that states: $$a^{\ln b} = b^{\ln a}$$. Thus, we rewrite the expression as: $$x^{1 /(1+\ln x)} = e^{\frac{\ln x}{1+\ln x}}$$ Now, we need to consider the behavior of this expression as x approaches 0 from the positive side.

Take the natural logarithm of both sides

Since the expression involves natural logarithms, let's take the natural logarithm of both sides: $$\ln\left(e^{\frac{\ln x}{1+\ln x}}\right) = \frac{\ln x}{1+\ln x}$$ Now, consider the limit as x approaches 0 from the positive side: $$\lim_{x \rightarrow 0^{+}}\frac{\ln x}{1+\ln x}$$

Apply L'Hôpital's rule

The current expression is now in a form suitable for applying L'Hôpital's rule since it is in the form of \(\frac{0}{0}\) as x approaches 0. We will now differentiate the numerator and denominator with respect to x: Numerator: $$\frac{d(\ln x)}{dx} = \frac{1}{x}$$ Denominator: $$\frac{d(1+\ln x)}{dx} = \frac{1}{x}$$ Now, apply L'Hôpital's rule: $$\lim_{x \rightarrow 0^{+}}\frac{\frac{1}{x}}{\frac{1}{x}}$$

Evaluate the limit

Simplify the expression and evaluate the limit: $$\lim_{x \rightarrow 0^{+}}\frac{\frac{1}{x}}{\frac{1}{x}} = \lim_{x \rightarrow 0^{+}} 1 = 1$$ Now that we've found the value of the limit of the expression after taking its natural logarithm, we must take the exponent of the result to find the limit of the original expression.

Find the limit of the original expression

Now, exponentiate the result: $$e^{\lim_{x \rightarrow 0^{+}} 1} = e^1 = e$$ So, the limit of the original expression is: $$\lim _{x \rightarrow 0^{+}} x^{1 /(1+\ln x)}= e$$

Key concepts

Natural Logarithms

Natural logarithms are a specific kind of logarithm where the base is the mathematical constant \( e \), approximately equal to 2.71828. The natural logarithm of a number \( x \) is usually denoted as \( \ln x \). This logarithm is particularly useful in calculus and mathematical analysis because it simplifies several complex computations involving exponential growth or decay.
Natural logarithms have some key properties that make them valuable:

• Logarithmic identity: \( \ln(e^x) = x \) for any real number \( x \).
• Product rule: \( \ln(ab) = \ln a + \ln b \).
• Quotient rule: \( \ln\left(\frac{a}{b}\right) = \ln a - \ln b \).
• Power rule: \( \ln(a^b) = b \ln a \).

In the solution of the limit exercise, taking the natural logarithm allowed us to rewrite the initial expression and simplify the calculation, making it easier to apply L'Hôpital's Rule.

Limits

In calculus, limits help us understand how a function behaves as it approaches a specific point. They give us insight into the continuity and possible discontinuities of functions. For instance, when calculating the limit \( \lim_{x \rightarrow 0^{+}} x^{1 /(1+\ln x)} \), we are determining the behavior of the expression as \( x \) nears zero from the positive side.
The concept of limits is foundational because it builds the groundwork for defining derivatives and integrals. The notation \( \lim_{x \to c} f(x) = L \) signifies that as \( x \) gets closer to \( c \), \( f(x) \) approaches \( L \).
In the exercise, transforming the problem into a limit of the form \( \lim_{x \rightarrow 0^{+}}\frac{\ln x}{1+\ln x} \) was essential for applying further calculus tools like L'Hôpital's Rule, which ultimately led us to solve the initial limit.

Differentiation

Differentiation is a critical tool in calculus used to find rates of change or slopes of curves. It calculates the derivative of a function, which reveals how the function itself changes relative to changes in its input. For example, if \( y = f(x) \), the derivative \( f'(x) \) tells us how \( y \) changes for a small change in \( x \).
In the context of L'Hôpital's Rule, differentiation is used to evaluate limits of indeterminate forms such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). When faced with these forms, differentiating the numerator and the denominator separately can provide a clearer path to finding the limit.
During the process of solving the given exercise, we applied differentiation to both the numerator \( \ln x \) and the denominator \( 1+\ln x \), obtaining derivatives \( \frac{1}{x} \) for both. This application of L'Hôpital's Rule simplified our evaluation of the limit, leading directly to a solution.