Question

Berechnen Sie die Grenzwerte $$ \lim _{x \rightarrow \infty} \frac{\ln x+\sqrt{x}}{x}, \quad \lim _{x \rightarrow 0} \frac{\sin \left(x^{3}\right)}{x^{2}}, \quad \lim _{x \rightarrow \frac{\pi}{2}} \frac{\cos x}{\frac{\pi}{2}-x}. $$

Short answer

The limits are 0, 0, and 1.

Step-by-step solution

Analyze the First Limit Problem

We need to find the limit \( \lim_{x \rightarrow \infty} \frac{\ln x + \sqrt{x}}{x} \). As \( x \to \infty \), both \( \ln x \) and \( \sqrt{x} \) grow, but \( \sqrt{x} \) grows faster than \( \ln x \). Therefore, the leading term is \( \sqrt{x} \). We rewrite:\[\frac{\ln x + \sqrt{x}}{x} = \frac{\ln x}{x} + \frac{\sqrt{x}}{x}\]Analyzing each term separately:- \( \frac{\ln x}{x} \to 0 \) because the logarithm grows slower than a linear function.- \( \frac{\sqrt{x}}{x} = \frac{1}{\sqrt{x}} \to 0 \) as \( x \to \infty \).Thus, \( \lim_{x \rightarrow \infty} \frac{\ln x + \sqrt{x}}{x} = 0 \).

Analyze the Second Limit Problem

Next, we find \( \lim_{x \rightarrow 0} \frac{\sin(x^3)}{x^2} \). As \( x \to 0 \), we can use the small angle approximation \( \sin y \approx y \) for \( y \to 0 \). Thus, \( \sin(x^3) \approx x^3 \). We substitute this approximation:\[\frac{\sin(x^3)}{x^2} \approx \frac{x^3}{x^2} = x\]As \( x \to 0 \), \( x \to 0 \).Therefore, \( \lim_{x \rightarrow 0} \frac{\sin(x^3)}{x^2} = 0 \).

Analyze the Third Limit Problem

We now find \( \lim_{x \rightarrow \frac{\pi}{2}} \frac{\cos x}{\frac{\pi}{2} - x} \). This is a \( \frac{0}{0} \) indeterminate form when \( x = \frac{\pi}{2} \), so we apply L'Hôpital's Rule:Take the derivative of the numerator and the denominator:- Derivative of \( \cos x = -\sin x \)- Derivative of \( \frac{\pi}{2} - x = -1 \)Applying L'Hôpital's Rule:\[\lim_{x \rightarrow \frac{\pi}{2}} \frac{\cos x}{\frac{\pi}{2} - x} = \lim_{x \rightarrow \frac{\pi}{2}} \frac{-\sin x}{-1} = \lim_{x \rightarrow \frac{\pi}{2}} \sin x\]As \( x \to \frac{\pi}{2} \), \( \sin x \to 1 \).Thus, \( \lim_{x \rightarrow \frac{\pi}{2}} \frac{\cos x}{\frac{\pi}{2} - x} = 1 \).

Key concepts

L'Hôpital's Rule

When you encounter limits that lead to indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), L'Hôpital's Rule is a handy technique. This rule helps in evaluating the limit by taking derivatives of the numerator and the denominator separately. Remember that L'Hôpital's Rule is only applicable when you start with an indeterminate form.

To use L'Hôpital's Rule:

• First, ensure the limit results in an indeterminate form.
• Then, take the derivative of the numerator and the derivative of the denominator.
• Finally, compute the limit of the new fraction.

In the context of the exercise, applying L'Hôpital's Rule to \( \lim_{x \rightarrow \frac{\pi}{2}} \frac{\cos x}{\frac{\pi}{2} - x} \) requires differentiating \( \cos x \) to get \( -\sin x \) and differentiating \( \frac{\pi}{2} - x \) to get \( -1 \). This simplifies the original problem, leading to the limit \( \lim_{x \rightarrow \frac{\pi}{2}} \sin x = 1 \).

Indeterminate Forms

Indeterminate forms often arise in calculus when you attempt to evaluate limits. They appear as ambiguous or undefined expressions, such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These forms signal the need for further analysis or transformation to find the actual limit.

When faced with indeterminate forms, you can often simplify them using algebra or calculus techniques like:

• Factoring and canceling terms.
• Rationalizing expressions.
• Applying L'Hôpital's Rule in applicable cases.

In our exercise, the limit \( \lim_{x \rightarrow \frac{\pi}{2}} \frac{\cos x}{\frac{\pi}{2} - x} \) results in a \( \frac{0}{0} \) form, indicating an opportunity to apply L'Hôpital's Rule. This helps redefine the limit into a solvable expression by differentiating the numerator and the denominator.

Small Angle Approximation

The small angle approximation is a useful tool in trigonometry and calculus. It is particularly handy for simplifying the expressions involving angles approaching zero. The key approximation is that \( \sin y \approx y \) when \( y \) is near zero. This simplification allows you to easily handle otherwise complex limits.

In the exercise, when evaluating \( \lim_{x \rightarrow 0} \frac{\sin(x^3)}{x^2} \), the small angle approximation \( \sin(x^3) \approx x^3 \) was used. This transforms the expression to \( \frac{x^3}{x^2} = x \), which as \( x \to 0 \), simplifies directly to 0. This approach shows how approximations can simplify the evaluation of limits.

Growth Rates of Functions

Understanding the growth rates of functions is crucial for comparing terms when evaluating limits. Different functions grow at different rates as \( x \to \infty \). For instance, logarithmic functions grow slower than polynomial functions, and polynomial functions grow slower than exponential functions.

When evaluating \( \lim_{x \rightarrow \infty} \frac{\ln x + \sqrt{x}}{x} \), acknowledging the growth rates allows simplification of the limit. Here, \( \ln x \) grows slower than both \( \sqrt{x} \) and \( x \), meaning it's negligible in comparison. Additionally, \( \sqrt{x} \) grows slower than \( x \), rendering its effect diminishable in the fraction. The main idea is to focus on the dominant term that dictates the behavior of the function as \( x \to \infty \), which in this case confirms the limit to be 0. Recognizing these growth rates offers clarity in comprehending how limits behave.