Question

$\lim_{x \to 2} \frac{3^x + 3^{3-x}-12}{3^{-x/2}-3^{1-x}}$ is equal to ____.

Answer 1

36

Answer 2

The problem asks us to evaluate the limit: $\lim_{x \to 2} \frac{3^x + 3^{3-x}-12}{3^{-x/2}-3^{1-x}}$

Step 1: Check the form of the limit. Substitute $x=2$ into the expression: Numerator: $3^2 + 3^{3-2} - 12 = 3^2 + 3^1 - 12 = 9 + 3 - 12 = 12 - 12 = 0$. Denominator: $3^{-2/2} - 3^{1-2} = 3^{-1} - 3^{-1} = \frac{1}{3} - \frac{1}{3} = 0$. Since the limit is of the $\frac{0}{0}$ indeterminate form, we can proceed with algebraic manipulation or L'Hopital's rule.

Step 2: Use substitution to simplify the expression. Let $y = 3^x$. As $x \to 2$, $y \to 3^2 = 9$. Now, express the terms in the limit in terms of $y$:

• $3^x = y$
• $3^{3-x} = 3^3 \cdot 3^{-x} = \frac{27}{3^x} = \frac{27}{y}$
• $3^{-x/2} = (3^x)^{-1/2} = y^{-1/2} = \frac{1}{\sqrt{y}}$
• $3^{1-x} = 3^1 \cdot 3^{-x} = \frac{3}{3^x} = \frac{3}{y}$

Substitute these into the original expression: $\lim_{y \to 9} \frac{y + \frac{27}{y} - 12}{\frac{1}{\sqrt{y}} - \frac{3}{y}}$

Step 3: Simplify the complex fraction. Combine terms in the numerator and denominator: Numerator: $y + \frac{27}{y} - 12 = \frac{y^2 + 27 - 12y}{y} = \frac{y^2 - 12y + 27}{y}$ Denominator: $\frac{1}{\sqrt{y}} - \frac{3}{y} = \frac{\sqrt{y}}{y} - \frac{3}{y} = \frac{\sqrt{y} - 3}{y}$

Now the limit expression becomes: $\lim_{y \to 9} \frac{\frac{y^2 - 12y + 27}{y}}{\frac{\sqrt{y} - 3}{y}}$ Since $y \to 9$, $y \neq 0$, so we can cancel $y$ from the numerator and denominator of the larger fraction: $\lim_{y \to 9} \frac{y^2 - 12y + 27}{\sqrt{y} - 3}$

Step 4: Factorize the numerator. The quadratic expression in the numerator is $y^2 - 12y + 27$. We need to find two numbers that multiply to 27 and add to -12. These numbers are -3 and -9. So, $y^2 - 12y + 27 = (y-3)(y-9)$.

Step 5: Factorize the term causing the indeterminate form. The term causing the $0/0$ form is $(\sqrt{y}-3)$ in the denominator. We can relate $y-9$ to this term using the difference of squares formula, $a^2 - b^2 = (a-b)(a+b)$. Here, $y-9 = (\sqrt{y})^2 - 3^2 = (\sqrt{y}-3)(\sqrt{y}+3)$.

Step 6: Substitute factored forms and cancel common terms. Substitute the factored forms back into the limit expression: $\lim_{y \to 9} \frac{(y-3)(y-9)}{\sqrt{y} - 3} = \lim_{y \to 9} \frac{(y-3)(\sqrt{y}-3)(\sqrt{y}+3)}{\sqrt{y} - 3}$ Since $y \to 9$, $y \neq 9$, which implies $\sqrt{y} \neq 3$. Therefore, we can cancel the common factor $(\sqrt{y}-3)$: $\lim_{y \to 9} (y-3)(\sqrt{y}+3)$

Step 7: Evaluate the limit. Now, substitute $y=9$ into the simplified expression: $(9-3)(\sqrt{9}+3) = (6)(3+3) = (6)(6) = 36$

The value of the limit is 36.