Answer:
[tex]\displaystyle \lim_{x \to 0^+} x^\big{\sqrt{x}} = 1[/tex]
General Formulas and Concepts:
Algebra II
- Natural logarithms ln and Euler's number e
- Logarithmic Property [Exponential]: [tex]\displaystyle log(a^b) = b \cdot log(a)[/tex]
Calculus
Limits
- Right-Side Limit: [tex]\displaystyle \lim_{x \to c^+} f(x)[/tex]
- Left-Side Limit: [tex]\displaystyle \lim_{x \to c^-} f(x)[/tex]
Limit Rule [Variable Direct Substitution]: [tex]\displaystyle \lim_{x \to c} x = c[/tex]
L’Hopital’s Rule: [tex]\displaystyle \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}[/tex]
Differentiation
- Derivatives
- Derivative Notation
Basic Power Rule:
- f(x) = cxⁿ
- f’(x) = c·nxⁿ⁻¹
Step-by-step explanation:
We are given the following limit:
[tex]\displaystyle \lim_{x \to 0^+} x^\big{\sqrt{x}}[/tex]
Substituting in x = 0 using the limit rule, we have an indeterminate form:
[tex]\displaystyle \lim_{x \to 0^+} x^\big{\sqrt{x}} = 0^0[/tex]
We need to rewrite this indeterminate form to another form to use L'Hopital's Rule. Let's set our limit as a function:
[tex]\displaystyle y = \lim_{x \to 0^+} x^\big{\sqrt{x}}[/tex]
Take the ln of both sides:
[tex]\displaystyle lny = ln \Big( \lim_{x \to 0^+} x^\big{\sqrt{x}} \Big)[/tex]
Rewrite the limit by including the ln in the inside:
[tex]\displaystyle lny = \lim_{x \to 0^+} ln \big( x^\big{\sqrt{x}} \big)[/tex]
Rewrite the limit once more using logarithmic properties:
[tex]\displaystyle lny = \lim_{x \to 0^+} \sqrt{x}ln(x)[/tex]
Rewrite the limit again:
[tex]\displaystyle lny = \lim_{x \to 0^+} \frac{ln(x)}{\frac{1}{\sqrt{x}}}[/tex]
Substitute in x = 0 again using the limit rule, we have an indeterminate form in which we can use L'Hopital's Rule:
[tex]\displaystyle \lim_{x \to 0^+} \frac{ln(x)}{\frac{1}{\sqrt{x}}} = \frac{\infty}{\infty}[/tex]
Apply L'Hopital's Rule:
[tex]\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}}}}[/tex]
Simplify:
[tex]\displaystyle \lim_{x \to 0^+} \frac{\frac{1}{x}}{\frac{-1}{2x^\big{\frac{3}{2}}}} = \lim_{x \to 0^+} -2\sqrt{x}[/tex]
Redefine the limit:
[tex]\displaystyle lny = \lim_{x \to 0^+} -2\sqrt{x}[/tex]
Substitute in x = 0 once more using the limit rule:
[tex]\displaystyle \lim_{x \to 0^+} -2\sqrt{x} = -2\sqrt{0}[/tex]
Evaluating it, we have:
[tex]\displaystyle \lim_{x \to 0^+} -2\sqrt{x} = 0[/tex]
Substitute in the limit value:
[tex]\displaystyle lny = 0[/tex]
e both sides:
[tex]\displaystyle e^\big{lny} = e^\big{0}[/tex]
Simplify:
[tex]\displaystyle y = 1[/tex]
And we have our final answer.
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Limits
