How to Find Limits of Complex Fractions Using Algebraic Techniques

Great question! This is a beautiful limit problem that combines exponential and logarithmic functions. Let's work through it together! 🌟

1. What We're Solving

We need to find: lim(x→0) (e^(x²) - ln(e + x²)) / x²

This is an indeterminate form that will require some clever algebraic work to evaluate.

2. The Approach

When we substitute x = 0 directly, we get (e⁰ - ln(e + 0)) / 0² = (1 - ln(e)) / 0 = (1 - 1) / 0 = 0/0, which is indeterminate!

Our strategy will be to use Taylor series expansions around x = 0. This is perfect because we're looking at the limit as x approaches 0, and Taylor series help us understand the behavior of functions near a specific point.

3. Step-by-Step Solution

Step 1: Let's expand e^(x²) using the Taylor series for e^u where u = x²

• e^u = 1 + u + u²/2! + u³/3! + ...
• So e^(x²) = 1 + x² + (x²)²/2! + (x²)³/3! + ...
• e^(x²) = 1 + x² + x⁴/2 + x⁶/6 + ...

Step 2: Now let's expand ln(e + x²)

• We can rewrite this as: ln(e + x²) = ln(e(1 + x²/e)) = ln(e) + ln(1 + x²/e) = 1 + ln(1 + x²/e)
• Using the Taylor series for ln(1 + v) where v = x²/e:
• ln(1 + v) = v - v²/2 + v³/3 - ...
• So ln(1 + x²/e) = x²/e - (x²/e)²/2 + (x²/e)³/3 - ...
• Therefore: ln(e + x²) = 1 + x²/e - x⁴/(2e²) + x⁶/(3e³) - ...

Step 3: Now let's substitute back into our original expression e^(x²) - ln(e + x²) = (1 + x² + x⁴/2 + ...) - (1 + x²/e - x⁴/(2e²) + ...)

Step 4: Simplify the numerator = 1 + x² + x⁴/2 + ... - 1 - x²/e + x⁴/(2e²) - ... = x² - x²/e + x⁴/2 + x⁴/(2e²) + ... = x²(1 - 1/e) + x⁴(1/2 + 1/(2e²)) + ...

Step 5: Divide by x² and take the limit [e^(x²) - ln(e + x²)] / x² = (1 - 1/e) + x²(1/2 + 1/(2e²)) + ...

As x → 0, all terms with positive powers of x vanish!

4. The Answer

lim(x→0) (e^(x²) - ln(e + x²)) / x² = 1 - 1/e = (e-1)/e

5. Memory Tip

When you see limits involving e^(something) and ln(something) as x → 0, think Taylor series! The key insight is that the dominant terms (lowest powers) determine the limit's value. Everything else becomes negligible as x approaches 0.

You did great tackling this challenging problem! The combination of exponential and logarithmic functions might look intimidating at first, but breaking it down with Taylor series makes it much more manageable. Keep practicing these techniques – they're incredibly powerful tools! 💪