Problem 64 - Calculus Early Transcendentals

Question

Evaluate the following limits or explain why they do not exist. Check your results by graphing. $$\lim _{x \rightarrow 0}\left(2^{a x}+x\right)^{1 / x}, \text { for a constant } a$$

Step-by-Step Solution

Answer

Answer: The limit as $$x$$ approaches 0 for the given function is $$e^{a\ln2 + 1}$$.

Step 1: Apply natural logarithm to both sides of the expression

Let $$L$$ be the limit we're trying to find: $$L = \lim _{x \rightarrow 0}\left(2^{a x}+x\right)^{1 / x}$$ Apply natural logarithm (ln) to both sides: $$\ln{L} = \lim _{x \rightarrow 0} \ln{\left(2^{a x}+x\right)^{1 / x}}$$ Use the property of logarithms that allows us to bring the exponent down: $$\ln{L} = \lim _{x \rightarrow 0} \frac{1}{x} \ln{\left(2^{a x}+x\right)}$$

Step 2: Apply L'Hôpital's Rule

Now, we can apply L'Hôpital's Rule to the limit above since it has the form 0/0 as $$x$$ approaches 0. To do this, we need to take the derivative of the numerator and denominator with respect to $$x$$: Numerator's derivative: $$\frac{d}{dx}\left[\ln{\left(2^{a x}+x\right)}\right]= \frac{a2^{ax}\ln2 + 1}{2^{ax} + x}$$ Denominator's derivative: $$\frac{d}{dx}\left[x\right]=1$$ Now, we can apply L'Hôpital's Rule to the limit: $$\ln{L} = \lim_{x\rightarrow 0} \frac{\frac{a2^{ax}\ln2 + 1}{2^{ax} + x}}{1}$$ As $$x$$ approaches 0, the limit becomes: $$\ln{L} = \frac{a\cdot 2^0 \cdot \ln2 + 1}{2^0 + 0} = a\ln2 + 1$$

Step 3: Solve for the original limit

Now, we need to find the original limit $$L$$ by eliminating the natural logarithm from both sides of the expression. We do this by taking the exponential of both sides: $$L = e^{\ln{L}}$$ $$L = e^{a\ln2 + 1}$$ This expression represents the limit we were looking for. Indeed, this expression exists for all constant values of $$a$$, so the limit does exist. $$\lim _{x \rightarrow 0}\left(2^{ax}+x\right)^{1 / x} = e^{a\ln2 + 1}$$

Key Concepts

Limits

When we talk about limits in calculus, we are essentially exploring what happens to a function as the input values approach a certain point. In this particular exercise, we're interested in what happens to the expression \( (2^{ax} + x)^{1/x} \) as \( x \) approaches 0.

This means we are trying to find the value that the expression gets closer to, as \( x \) gets infinitesimally small. Limits are a fundamental aspect of calculus because they help define other concepts such as continuity, derivatives, and integrals, which are all pivotal to understanding complex mathematical ideas.

Approaching limits can be challenging due to indeterminate forms such as \( 0/0 \). Indeterminate forms do not have a basis for comparison and require further analysis or methods to evaluate, one of which is L'Hôpital's Rule.

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool in calculus used to evaluate limits that produce indeterminate forms like \( 0/0 \) or \( \infty/\infty \). When a limit results in one of these forms, direct substitution isn't enough, and special methods are needed.

The rule states that for functions \( f(x) \) and \( g(x) \) both approaching 0 (or \( \infty \)), \[ \lim_{x\to c} \frac{f(x)}{g(x)} = \lim_{x\to c} \frac{f'(x)}{g'(x)} \] if the limit on the right side exists. Here, \( f'(x) \) and \( g'(x) \) are the derivatives of the numerator and denominator, respectively.

In our calculation, after obtaining the form \( \frac{1}{x} \ln(2^{ax} + x) \), we encountered \( 0/0 \) as \( x \to 0 \), making it a prime candidate for L'Hôpital's Rule. By calculating the derivatives, we simplified the evaluation of this challenging limit.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is the logarithm with base \( e \), where \( e \approx 2.71828 \), a fundamental constant in mathematics.

In this exercise, applying the natural logarithm helps simplify the limit evaluation process. Specifically, logarithms have the useful property of turning exponents into multipliers, which makes it easier to work with expressions like \( (2^{ax}+x)^{1/x} \).

Applying \( \ln \) to the expression allows us to bring down the exponent, resulting in \( \ln L = \lim_{x \to 0} \frac{1}{x} \ln(2^{ax} + x) \). This transformation is crucial for simplifying our limit expression and serves as an intermediate step that is manageable to tackle with further calculus techniques like L'Hôpital's Rule.

Exponentiation

Exponentiation is the operation of raising one number to the power of another, as seen in terms like \( 2^{ax} \). This operation is a core component of many mathematical models and problems.

In our exercise, exponentiation plays a key role, intersecting with functions, logarithms, and calculus concepts. The expression \( (2^{ax} + x)^{1/x} \) involves both exponentiation and the limit process, revealing the depth of interest in how exponential models behave as they approach extreme values, like 0.

To solve our problem, we initially handle the exponent by converting our initial expression using logarithms. Exponents, when worked through with differentiation and logarithmic manipulation, reveal key insights about the smooth behavior or tendencies of the expression as \( x \) nears 0, which is made concrete when resolved through exponentiation back to the original power form.