Math Problem Statement

Find the limit. Use l'Hôpital's Rule where appropriate. If there is a more elementary method, consider using it. 1. \( \lim_{{x \to 0^+}} \sqrt{x} \ln x \) 2. \( \lim_{{x \to \infty}} x^2 e^x \) 3. \( \lim_{{x \to \infty}} \sqrt{x^2 + x} - x \) 4. \( \lim_{{x \to \pi/2^+}} (\sec x - \cot x) \) 5. \( \lim_{{x \to \infty}} (x^{1/x}) \) 6. \( \lim_{{x \to \infty}} \left( \frac{x}{x + 2} \right)^x \)

Solution

Let's solve each limit one by one.

Problem 1:

lim⁡x→0+xln⁡(x)\lim_{{x \to 0^+}} \sqrt{x} \ln(x)

Solution:

Rewrite x\sqrt{x} as x1/2x^{1/2}, so: lim⁡x→0+x1/2ln⁡(x)=lim⁡x→0+ln⁡(x)x−1/2\lim_{{x \to 0^+}} x^{1/2} \ln(x) = \lim_{{x \to 0^+}} \frac{\ln(x)}{x^{-1/2}} This is an indeterminate form of type −∞∞\frac{-\infty}{\infty}, so we can apply L'Hôpital's Rule: =lim⁡x→0+ddx(ln⁡(x))ddx(x−1/2)=lim⁡x→0+1x−12x−3/2= \lim_{{x \to 0^+}} \frac{\frac{d}{dx}(\ln(x))}{\frac{d}{dx}(x^{-1/2})} = \lim_{{x \to 0^+}} \frac{\frac{1}{x}}{-\frac{1}{2}x^{-3/2}} Simplify: =lim⁡x→0+−2x=−∞= \lim_{{x \to 0^+}} \frac{-2}{\sqrt{x}} = -\infty

So, [ \lim_{{

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Math Problem Analysis

Mathematical Concepts

Limits
Indeterminate Forms
L'Hôpital's Rule

Formulas

\( \lim_{{x \to a}} \frac{f(x)}{g(x)} \) using L'Hôpital's Rule if \( \frac{0}{0} \) or \( \frac{\pm \infty}{\pm \infty} \) form
Exponential limits \( \lim_{{x \to \infty}} \left( 1 + \frac{1}{x} \right)^x = e \)

Theorems

L'Hôpital's Rule
Limit laws
Indeterminate forms

Suitable Grade Level

College Calculus

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