Question

Find the limit of the formula given​

Answer 1

Answer:

$\displaystyle \lim_{x \to 0^+} x^\big{\sqrt{x}} = 1$

General Formulas and Concepts:

Algebra II

• Natural logarithms ln and Euler's number e
• Logarithmic Property [Exponential]:                                                             $\displaystyle log(a^b) = b \cdot log(a)$

Calculus

Limits

• Right-Side Limit:                                                                                             $\displaystyle \lim_{x \to c^+} f(x)$
• Left-Side Limit:                                                                                               $\displaystyle \lim_{x \to c^-} f(x)$

Limit Rule [Variable Direct Substitution]:                                                             $\displaystyle \lim_{x \to c} x = c$

L’Hopital’s Rule:                                                                                                     $\displaystyle \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$

Differentiation

• Derivatives
• Derivative Notation

Basic Power Rule:

1. f(x) = cxⁿ
2. f’(x) = c·nxⁿ⁻¹  

Step-by-step explanation:

We are given the following limit:

$\displaystyle \lim_{x \to 0^+} x^\big{\sqrt{x}}$

Substituting in x = 0 using the limit rule, we have an indeterminate form:

$\displaystyle \lim_{x \to 0^+} x^\big{\sqrt{x}} = 0^0$

We need to rewrite this indeterminate form to another form to use L'Hopital's Rule. Let's set our limit as a function:

$\displaystyle y = \lim_{x \to 0^+} x^\big{\sqrt{x}}$

Take the ln of both sides:

$\displaystyle lny = ln \Big( \lim_{x \to 0^+} x^\big{\sqrt{x}} \Big)$

Rewrite the limit by including the ln in the inside:

$\displaystyle lny = \lim_{x \to 0^+} ln \big( x^\big{\sqrt{x}} \big)$

Rewrite the limit once more using logarithmic properties:

$\displaystyle lny = \lim_{x \to 0^+} \sqrt{x}ln(x)$

Rewrite the limit again:

$\displaystyle lny = \lim_{x \to 0^+} \frac{ln(x)}{\frac{1}{\sqrt{x}}}$

Substitute in x = 0 again using the limit rule, we have an indeterminate form in which we can use L'Hopital's Rule:

$\displaystyle \lim_{x \to 0^+} \frac{ln(x)}{\frac{1}{\sqrt{x}}} = \frac{\infty}{\infty}$

Apply L'Hopital's Rule:

$\displaystyle \lim_{x \to 0^+} \frac{ln(x)}{\frac{1}{\sqrt{x}}} = \lim_{x \to 0^+} \frac{\frac{1}{x}}{\frac{-1}{2x^\big{\frac{3}{2}}}}$

Simplify:

$\displaystyle \lim_{x \to 0^+} \frac{\frac{1}{x}}{\frac{-1}{2x^\big{\frac{3}{2}}}} = \lim_{x \to 0^+} -2\sqrt{x}$

Redefine the limit:

$\displaystyle lny = \lim_{x \to 0^+} -2\sqrt{x}$

Substitute in x = 0 once more using the limit rule:

$\displaystyle \lim_{x \to 0^+} -2\sqrt{x} = -2\sqrt{0}$

Evaluating it, we have:

$\displaystyle \lim_{x \to 0^+} -2\sqrt{x} = 0$

Substitute in the limit value:

$\displaystyle lny = 0$

e both sides:

$\displaystyle e^\big{lny} = e^\big{0}$

Simplify:

$\displaystyle y = 1$

And we have our final answer.

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit:  Limits