Answer:
[tex]\displaystyle \lim_{h \to 0} \frac{\sqrt{3} \Delta (\frac{\pi}{3})}{h^2} = \boxed{ 144 \sqrt{3} }[/tex]
General Formulas and Concepts:
Pre-Calculus
2x2 Matrix Determinant:
[tex]\displaystyle \left| \begin{array}{ccc} a & b \\ c & d \end{array} \right| = ad - bc[/tex]
3x3 Matrix Determinant:
[tex]\displaystyle \left| \begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \end{array} \right| = a \left| \begin{array}{ccc} e & f \\ h & i \end{array} \right| - b \left| \begin{array}{ccc} d & f \\ g & i \end{array} \right| + c \left| \begin{array}{ccc} d & e \\ g & h \end{array} \right|[/tex]
Calculus
Limits
Limit Rule [Variable Direct Substitution]:
[tex]\displaystyle \lim_{x \to c} x = c[/tex]
Limit Property [Multiplied Constant]:
[tex]\displaystyle \lim_{x \to c} bf(x) = b \lim_{x \to c} f(x)[/tex]
Special Limit Rule [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]
Derivatives
- Derivatives
- Derivative Notation
Derivative Property [Addition/Subtraction]:
[tex]\displaystyle (u + v)' = u' + v'[/tex]
Derivative Rule [Chain Rule]:
[tex]\displaystyle [u(v)]' = u'(v)v'[/tex]
Step-by-step explanation:
*Note:
I will not be able to fit in all the derivative work and will assume you can take derivatives with ease.
Step 1: Define
Identify given.
[tex]\displaystyle \Delta (x) = \left| \begin{array}{ccc} \tan x & \tan (x + h) & \tan (x + 2h) \\ \tan (x + 2h) & \tan x & \tan (x + h) \\ \tan (x + h) & \tan (x + 2h) & \tan x \end{array} \right|[/tex]
[tex]\displaystyle \lim_{h \to 0} \frac{\sqrt{3} \Delta (\frac{\pi}{3})}{h^2}[/tex]
Step 2: Find Limit Pt. 1
- [Function] Simplify [3x3 and 2x2 Matrix Determinant]:
[tex]\displaystyle \Delta (x) = \tan^3 (2h + x) + \tan^3 (h + x) + \tan^3 x - 3 \tan x \tan (h + x) \tan (2h + x)[/tex] - [Function] Substitute in x:
[tex]\displaystyle \Delta \bigg( \frac{\pi}{3} \bigg) = \tan^3 \bigg( 2h+ \frac{\pi}{3} \bigg) + \tan^3 \bigg( h + \frac{\pi}{3} \bigg) + 3\sqrt{3} - 3\sqrt{3} \tan \bigg( h + \frac{\pi}{3} \bigg) \tan \bigg( 2h+ \frac{\pi}{3} \bigg)[/tex]
Step 3: Find Limit Pt. 2
- [Limit] Rewrite [Limit Property - Multiplied Constant]:
[tex]\displaystyle \lim_{h \to 0} \frac{\sqrt{3} \Delta (\frac{\pi}{3})}{h^2} = \sqrt{3} \lim_{h \to 0} \frac{\Delta (\frac{\pi}{3})}{h^2}[/tex] - [Limit] Apply Limit Rule [Variable Direct Substitution]:
[tex]\displaystyle \lim_{h \to 0} \frac{\sqrt{3} \Delta (\frac{\pi}{3})}{h^2} = \sqrt{3} \bigg( \frac{0}{0} \bigg)[/tex]
Since we have an indeterminant form, we will have to use L'Hopital's Rule. We can differentiate using basic differentiation techniques listed above under "Calculus":
[tex]\displaystyle \frac{d \Delta (\frac{\pi}{3})}{dh} = -3\sqrt{3} \bigg[ \tan^2 \bigg( h + \frac{\pi}{3} \bigg) + 1 \bigg] \tan \bigg( 2h + \frac{\pi}{3} \bigg) + tan^2 \bigg( h + \frac{\pi}{3} \bigg) \bigg[ 3 \tan^2 \bigg( h + \frac{\pi}{3} + 3 \bigg] - 3\sqrt{3} \tan \bigg( h + \frac{\pi}{3} \bigg) \bigg[ 2 \tan^2 \bigg( 2h + \frac{\pi}{3} \bigg) + 2 \bigg] + \tan^2 \bigg( 2h + \frac{\pi}{3} \bigg) \bigg[ 6 \tan^2 \bigg( 2h + \frac{\pi}{3} \bigg) + 6 \bigg][/tex]
[tex]\displaystyle \frac{d}{dh} h^2 = 2h[/tex]
Using L'Hopital's Rule, we can substitute the derivatives and evaluate again. When we do so, we should get another indeterminant form. We will need to use L'Hopital's Rule again:
[tex]\displaystyle \frac{d^2 \Delta (\frac{\pi}{3})}{dh^2} = \tan \bigg( h + \frac{\pi}{3} \bigg) \bigg[ 2 \tan^2 \bigg( h + \frac{\pi}{3} \bigg) + 2 \bigg] \bigg[ \tan^2 \bigg( h + \frac{\pi}{3} \bigg) + 1 \bigg] - 2\sqrt{3} \bigg[ \tan^2 \bigg( h + \frac{\pi}{3} \bigg) + 1 \bigg] \bigg[ \tan^2 \bigg( 2h + \frac{\pi}{3} \bigg) + 1 \bigg] - \sqrt{3} \bigg[ \tan^2 \bigg( h + \frac{\pi}{3} \bigg) + 1 \bigg] \bigg[ 2 \tan^2 \bigg( 2h + \frac{\pi}{3} \bigg) + 2 \bigg][/tex]
[tex]\displaystyle + \tan^3 \bigg( h + \frac{\pi}{3} \bigg) \bigg[ 2 \tan^2 \bigg( h + \frac{\pi}{3} \bigg) + 2 \bigg] - \sqrt{3} \tan \bigg( h + \frac{\pi}{3} \bigg) \tan \bigg( 2h + \frac{\pi}{3} \bigg) \bigg[ 2 \tan^2 \bigg( h + \frac{\pi}{3} \bigg) + 2 \bigg] + \tan \bigg( 2h + \frac{\pi}{3} \bigg) \bigg[ 2 \tan^2 \bigg( 2h + \frac{\pi}{3} \bigg) + 2 \bigg] \bigg[ 4 \tan^2 \bigg( 2h + \frac{\pi}{3} \bigg) + 4 \bigg][/tex]
[tex]\displaystyle - 2\sqrt{3} \tan \bigg( h + \frac{\pi}{3} \bigg) \tan \bigg( 2h + \frac{\pi}{3} \bigg) \bigg[ 4 \tan^2 \bigg( 2h + \frac{\pi}{3} \bigg) + 4 \bigg] + 2 \tan^3 \bigg( 2h + \frac{\pi}{3} \bigg) \bigg[ 4 \tan^2 \bigg( 2h + \frac{\pi}{3} \bigg) + 4 \bigg][/tex]
[tex]\displaystyle \frac{d^2}{dh^2} h^2 = 2[/tex]
Substituting in the 2nd derivative found via L'Hopital's Rule should now give us a numerical value when evaluating the limit using limit rules and the unit circle:
[tex]\displaystyle \lim_{h \to 0} \frac{\sqrt{3} \Delta (\frac{\pi}{3})}{h^2} = \boxed{ 144 \sqrt{3} }[/tex]
∴ we have evaluated the given limit.
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