Question

Please help! Related to limits! 100 points!

Answer 1

Answer:

$\displaystyle \lim_{h \to 0} \frac{\sqrt{3} \Delta (\frac{\pi}{3})}{h^2} = \boxed{ 144 \sqrt{3} }$

General Formulas and Concepts:
Pre-Calculus

2x2 Matrix Determinant:
$\displaystyle \left| \begin{array}{ccc} a & b \\ c & d \end{array} \right| = ad - bc$

3x3 Matrix Determinant:
$\displaystyle \left| \begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \end{array} \right| = a \left| \begin{array}{ccc} e & f \\ h & i \end{array} \right| - b \left| \begin{array}{ccc} d & f \\ g & i \end{array} \right| + c \left| \begin{array}{ccc} d & e \\ g & h \end{array} \right|$

Calculus

Limits

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

Limit Property [Multiplied Constant]:
$\displaystyle \lim_{x \to c} bf(x) = b \lim_{x \to c} f(x)$

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

Derivatives

• Derivatives
• Derivative Notation

Derivative Property [Addition/Subtraction]:
$\displaystyle (u + v)' = u' + v'$

Derivative Rule [Chain Rule]:
$\displaystyle [u(v)]' = u'(v)v'$

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.

$\displaystyle \Delta (x) = \left| \begin{array}{ccc} \tan x & \tan (x + h) & \tan (x + 2h) \\ \tan (x + 2h) & \tan x & \tan (x + h) \\ \tan (x + h) & \tan (x + 2h) & \tan x \end{array} \right|$

$\displaystyle \lim_{h \to 0} \frac{\sqrt{3} \Delta (\frac{\pi}{3})}{h^2}$

Step 2: Find Limit Pt. 1

1. [Function] Simplify [3x3 and 2x2 Matrix Determinant]:
   $\displaystyle \Delta (x) = \tan^3 (2h + x) + \tan^3 (h + x) + \tan^3 x - 3 \tan x \tan (h + x) \tan (2h + x)$
2. [Function] Substitute in x:
   $\displaystyle \Delta \bigg( \frac{\pi}{3} \bigg) = \tan^3 \bigg( 2h+ \frac{\pi}{3} \bigg) + \tan^3 \bigg( h + \frac{\pi}{3} \bigg) + 3\sqrt{3} - 3\sqrt{3} \tan \bigg( h + \frac{\pi}{3} \bigg) \tan \bigg( 2h+ \frac{\pi}{3} \bigg)$

Step 3: Find Limit Pt. 2

1. [Limit] Rewrite [Limit Property - Multiplied Constant]:
   $\displaystyle \lim_{h \to 0} \frac{\sqrt{3} \Delta (\frac{\pi}{3})}{h^2} = \sqrt{3} \lim_{h \to 0} \frac{\Delta (\frac{\pi}{3})}{h^2}$
2. [Limit] Apply Limit Rule [Variable Direct Substitution]:
   $\displaystyle \lim_{h \to 0} \frac{\sqrt{3} \Delta (\frac{\pi}{3})}{h^2} = \sqrt{3} \bigg( \frac{0}{0} \bigg)$

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":

$\displaystyle \frac{d \Delta (\frac{\pi}{3})}{dh} = -3\sqrt{3} \bigg[ \tan^2 \bigg( h + \frac{\pi}{3} \bigg) + 1 \bigg] \tan \bigg( 2h + \frac{\pi}{3} \bigg) + tan^2 \bigg( h + \frac{\pi}{3} \bigg) \bigg[ 3 \tan^2 \bigg( h + \frac{\pi}{3} + 3 \bigg] - 3\sqrt{3} \tan \bigg( h + \frac{\pi}{3} \bigg) \bigg[ 2 \tan^2 \bigg( 2h + \frac{\pi}{3} \bigg) + 2 \bigg] + \tan^2 \bigg( 2h + \frac{\pi}{3} \bigg) \bigg[ 6 \tan^2 \bigg( 2h + \frac{\pi}{3} \bigg) + 6 \bigg]$

$\displaystyle \frac{d}{dh} h^2 = 2h$

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:

$\displaystyle \frac{d^2 \Delta (\frac{\pi}{3})}{dh^2} = \tan \bigg( h + \frac{\pi}{3} \bigg) \bigg[ 2 \tan^2 \bigg( h + \frac{\pi}{3} \bigg) + 2 \bigg] \bigg[ \tan^2 \bigg( h + \frac{\pi}{3} \bigg) + 1 \bigg] - 2\sqrt{3} \bigg[ \tan^2 \bigg( h + \frac{\pi}{3} \bigg) + 1 \bigg] \bigg[ \tan^2 \bigg( 2h + \frac{\pi}{3} \bigg) + 1 \bigg] - \sqrt{3} \bigg[ \tan^2 \bigg( h + \frac{\pi}{3} \bigg) + 1 \bigg] \bigg[ 2 \tan^2 \bigg( 2h + \frac{\pi}{3} \bigg) + 2 \bigg]$

$\displaystyle + \tan^3 \bigg( h + \frac{\pi}{3} \bigg) \bigg[ 2 \tan^2 \bigg( h + \frac{\pi}{3} \bigg) + 2 \bigg] - \sqrt{3} \tan \bigg( h + \frac{\pi}{3} \bigg) \tan \bigg( 2h + \frac{\pi}{3} \bigg) \bigg[ 2 \tan^2 \bigg( h + \frac{\pi}{3} \bigg) + 2 \bigg] + \tan \bigg( 2h + \frac{\pi}{3} \bigg) \bigg[ 2 \tan^2 \bigg( 2h + \frac{\pi}{3} \bigg) + 2 \bigg] \bigg[ 4 \tan^2 \bigg( 2h + \frac{\pi}{3} \bigg) + 4 \bigg]$

$\displaystyle - 2\sqrt{3} \tan \bigg( h + \frac{\pi}{3} \bigg) \tan \bigg( 2h + \frac{\pi}{3} \bigg) \bigg[ 4 \tan^2 \bigg( 2h + \frac{\pi}{3} \bigg) + 4 \bigg] + 2 \tan^3 \bigg( 2h + \frac{\pi}{3} \bigg) \bigg[ 4 \tan^2 \bigg( 2h + \frac{\pi}{3} \bigg) + 4 \bigg]$

$\displaystyle \frac{d^2}{dh^2} h^2 = 2$

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:

$\displaystyle \lim_{h \to 0} \frac{\sqrt{3} \Delta (\frac{\pi}{3})}{h^2} = \boxed{ 144 \sqrt{3} }$

∴ we have evaluated the given limit.

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