Problem 42 - Calculus for Biology and Medicine

Question

Use l'Hospital's rule to find the limits. $$ \lim _{x \rightarrow 0^{+}} x^{\sin x} $$

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Rewrite the expression

We start by rewriting the original limit expression in a more manageable form. We note that the power is a function of a variable, which suggests using the logarithmic transformation. We can express the given expression as follows:\[ x^{\sin x} = e^{(\ln(x^{\sin x}))} = e^{(\sin x \cdot \ln x)}. \]Now, our task is to find the limit of \( e^{(\sin x \cdot \ln x)} \) as \( x \to 0^+ \).

Step 2: Analyze \( \sin x \cdot \ln x \)

We focus on finding \( \lim_{x \rightarrow 0^{+}} (\sin x \cdot \ln x) \). As \( x \to 0^+ \), \( \ln x \to -\infty \) and \( \sin x \to 0 \). This leads to the indeterminate form \( 0 \times (-\infty) \). We rewrite it as:\[ \lim_{x \to 0^+} \frac{\ln x}{1/(\sin x)}. \]

Step 3: Apply l'Hospital's Rule

The expression \( \frac{\ln x}{1/(\sin x)} \) is in the \( \frac{-\infty}{\infty} \) form. We can now apply l'Hospital's Rule, which allows us to take the derivative of the numerator and denominator:- Derivative of the numerator: \( \frac{d}{dx}(\ln x) = \frac{1}{x} \).- Derivative of the denominator: \( \frac{d}{dx}(1/\sin x) = -\frac{\cos x}{\sin^2 x} = -\cot x \csc x \).Therefore, applying l'Hospital's Rule gives us:\[ \lim_{x \to 0^+} \frac{1/x}{-\cot x \csc x} = \lim_{x \to 0^{+}} \frac{-\sin x \cos x}{x} = \lim_{x \to 0^+} -\frac{1}{x} \cdot \sin x \cdot \cos x.\]

Step 4: Evaluate the simplified limit

We now simplify \( -\frac{1}{x} \cdot \sin x \cdot \cos x \) and find its limit as \( x \to 0^+ \). Recognizing that \( \sin x \sin x \) can be approximated as \( x \) when \( x \to 0 \), we have:\[ -\frac{x}{x} \cdot \cos x = -\cos x. \]Since \( \cos x \to 1 \) as \( x \to 0^+ \), the limit evaluates to \( 0 \). This means:\[ \lim_{x \to 0^+} (\sin x \cdot \ln x) = 0. \]

Step 5: Final limit evaluation

Since we found that \( \lim_{x \to 0^+} (\sin x \cdot \ln x) = 0 \), it follows:\[ \lim_{x \to 0^+} e^{(\sin x \cdot \ln x)} = e^0 = 1. \]

Key Concepts

Logarithmic Transformation

When faced with an expression where the exponent itself involves a variable, such as the original task of finding \( \lim_{x \to 0^+} x^{\sin x} \), using a logarithmic transformation can simplify the process.
Logarithmic transformations help convert these complex exponential expressions into products that are easier to analyze.

For instance, the expression \( x^{\sin x} \) can be rewritten as \( e^{(\ln(x^{\sin x}))} \). By applying the properties of logarithms, this becomes \( e^{(\sin x \cdot \ln x)} \).
This step turns our original expression into one that focuses on the exponent, which is crucial for further evaluation.

The beauty of this transformation is in its ability to handle exponentials in limits, paving the way for easier application of limit-solving techniques, like l'Hospital's Rule.

Indeterminate Forms

In calculus, an indeterminate form is a type of limit expression that does not allow immediate evaluation. It requires further simplification or manipulation to find a meaningful limit.
In our example, as \( x \to 0^+ \), the expression \( \sin x \cdot \ln x \) becomes an indeterminate form: \( 0 \times (-\infty) \).

This is because \( \ln x \to -\infty \) while \( \sin x \to 0 \), making direct computation impossible without additional steps.
To address this, we rewrite the expression as a quotient \( \frac{\ln x}{1/\sin x} \), transforming it into the form \( \frac{-\infty}{\infty} \).

Through this transformation, standard calculus techniques, such as l'Hospital's Rule, can be applied to find the limit efficiently.

Limit Evaluation

The final step in solving the original problem involves evaluating the limit using l'Hospital's Rule.
Once we rewrite the indeterminate form \( \frac{-\infty}{\infty} \), we apply l'Hospital's, which involves taking derivatives of the numerator and the denominator.

For \( \frac{\ln x}{1/\sin x} \), l'Hospital's Rule states to differentiate the numerator, \( \frac{d}{dx}(\ln x) = \frac{1}{x} \), and the denominator, which through derivative becomes \( -\cot x \cdot \csc x \).
By differentiating and simplifying, we get \( \lim_{x \to 0^+} -\frac{1}{x} \cdot \sin x \cdot \cos x \).

Recognizing that \( \sin x/x \approx 1 \) when \( x \to 0 \), the expression simplifies to \(-\cos x\), approaching 1, making the limit \( \lim_{x \to 0^+} e^{(\sin x \cdot \ln x)} = 1 \). Here, understanding each step is key to correctly applying these calculus techniques.