Math Problem Statement

Evaluate the limit: lim_{x→2} (5^x + 5^(2-x) - 26) / (5^(x/2) - 5^(1-x)).

Solution

The problem is to evaluate the limit:

lim⁡x→25x+52−x−265x/2−51−x\lim_{x \to 2} \frac{5^x + 5^{2-x} - 26}{5^{x/2} - 5^{1-x}}

Step 1: Substitute x=2x = 2 directly

Substituting x=2x = 2:

  • Numerator: 5x+52−x−26=52+52−2−26=25+1−26=05^x + 5^{2-x} - 26 = 5^2 + 5^{2-2} - 26 = 25 + 1 - 26 = 0
  • Denominator: 5x/2−51−x=52/2−51−2=51−5−1=5−15=255−15=2455^{x/2} - 5^{1-x} = 5^{2/2} - 5^{1-2} = 5^1 - 5^{-1} = 5 - \frac{1}{5} = \frac{25}{5} - \frac{1}{5} = \frac{24}{5}

Thus, the limit becomes: 0245=0\frac{0}{\frac{24}{5}} = 0

However, because the numerator and denominator both simplify to zero, we need to apply L'Hôpital's Rule to evaluate the limit.

Step 2: Apply L'Hôpital's Rule

Take the derivatives of the numerator and denominator.

Numerator:

= \frac{d}{dx} (5^x) + \frac{d}{dx}(5^{2-x}) - \frac{d}{dx}(26)$$ Using the derivative rule for $$a^x$$: $$\frac{d}{dx} a^x = a^x \ln(a)$$, we get: $$\frac{d}{dx} (5^x) = 5^x \ln(5)$$ $$\frac{d}{dx}(5^{2-x}) = -5^{2-x} \ln(5)$$ $$\frac{d}{dx}(26) = 0$$ So, the derivative of the numerator is: $$5^x \ln(5) - 5^{2-x} \ln(5)$$ #### Denominator: $$\frac{d}{dx} \left( 5^{x/2} - 5^{1-x} \right) = \frac{d}{dx}(5^{x/2}) - \frac{d}{dx}(5^{1-x})$$ For $$5^{x/2}$$, using the chain rule: $$\frac{d}{dx}(5^{x/2}) = 5^{x/2} \ln(5) \cdot \frac{1}{2}$$ For $$5^{1-x}$$: $$\frac{d}{dx}(5^{1-x}) = -5^{1-x} \ln(5)$$ Thus, the derivative of the denominator is: $$\frac{1}{2} 5^{x/2} \ln(5) + 5^{1-x} \ln(5)$$ --- ### Step 3: Simplify the limit The limit now becomes: $$\lim_{x \to 2} \frac{5^x \ln(5) - 5^{2-x} \ln(5)}{\frac{1}{2} 5^{x/2} \ln(5) + 5^{1-x} \ln(5)}$$ Cancel out $$\ln(5)$$ (non-zero) from numerator and denominator: $$\lim_{x \to 2} \frac{5^x - 5^{2-x}}{\frac{1}{2} 5^{x/2} + 5^{1-x}}$$ Substitute $$x = 2$$ again: - Numerator: $$5^x - 5^{2-x} = 5^2 - 5^{2-2} = 25 - 1 = 24$$ - Denominator: $$\frac{1}{2} 5^{x/2} + 5^{1-x} = \frac{1}{2} 5^{2/2} + 5^{1-2} = \frac{1}{2}(5) + \frac{1}{5} = \frac{5}{2} + \frac{1}{5}$$ Combine terms: $$\frac{5}{2} + \frac{1}{5} = \frac{25}{10} + \frac{2}{10} = \frac{27}{10}$$ Thus, the limit becomes: $$\frac{24}{\frac{27}{10}} = 24 \cdot \frac{10}{27} = \frac{240}{27} = \frac{80}{9}$$ --- ### Final Answer: The limit evaluates to: $$\boxed{\frac{80}{9}}$$ Do you have questions or need further clarification? --- ### Related Questions: 1. How do you use L’Hôpital’s Rule when both numerator and denominator are zero? 2. Can you explain the derivative of $$a^x$$ and why $$\ln(a)$$ appears? 3. How do you simplify fractions with different denominators? 4. How does exponential growth affect limits in calculus? 5. What are common cases that require L’Hôpital’s Rule? ### Tip: Always check if substitution works before applying L’Hôpital’s Rule to save time!

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Math Problem Analysis

Mathematical Concepts

Limits
Exponential Functions
L'Hôpital's Rule

Formulas

Derivative of a^x: d/dx(a^x) = a^x * ln(a)
L'Hôpital's Rule: lim(x→c) f(x)/g(x) = lim(x→c) f'(x)/g'(x) if f(c) = g(c) = 0 or ±∞
Simplifying fractional expressions

Theorems

L'Hôpital's Rule

Suitable Grade Level

Grades 11-12 or Calculus Level

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