How to Calculate Limits as x Approaches 0 in Calculus

Hello! This is a great limit problem that will help you practice some key calculus techniques. Let's work through it together!

What We're Solving:

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

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 an indeterminate form!

This means we need a special technique. Since we have a 0/0 form, we can use Taylor series expansions around x = 0. This is perfect because Taylor series help us understand the behavior of functions near a specific point by breaking them into polynomial pieces.

Step-by-Step Solution:

Step 1: Let's find the Taylor expansion of e^(x²) around x = 0 We know that e^u = 1 + u + u²/2! + u³/3! + ... Substituting u = x²: e^(x²) = 1 + x² + (x²)²/2! + ... = 1 + x² + x⁴/2 + ...

Step 2: Now let's find the Taylor expansion of 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 expansion ln(1 + v) = v - v²/2 + v³/3 + ... where v = x²/e: ln(1 + x²/e) = x²/e - (x²/e)²/2 + ... = x²/e - x⁴/(2e²) + ...

So: ln(e + x²) = 1 + x²/e - x⁴/(2e²) + ...

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

Step 4: 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, leaving us with: lim(x→0) = 1 - 1/e

The Answer:

1 - 1/e (which is approximately 0.632)

Memory Tip:

When you see a 0/0 indeterminate form with exponentials and logarithms, think "Taylor series time!" The key insight is that Taylor expansions let you "zoom in" on how functions behave near a point, turning complicated expressions into simple polynomials that are much easier to work with.

Great work tackling this challenging limit! The combination of recognizing the indeterminate form and knowing to use Taylor series shows real mathematical maturity. Keep practicing these techniques! 🌟