91Ó°ÊÓ

In calculus, the binomial approximation is a useful tool for simplifying expressions that involve roots and small values. When working with expressions like \( \sqrt{1 + t} \), especially as \( t \) approaches zero, you can approximate it using a binomial expansion.To understand this, think of the binomial theorem, which allows you to expand powers of binomials \((a + b)^n\). For small values of \( t \), the first few terms of this expansion are often enough to provide an accurate approximation. For example, the square root \( \sqrt{1+t} \) can be approximated by:

• \(1 + \frac{t}{2}\)
• minus higher order terms like \(\frac{t^2}{8}\), since they become very small as \( t \) goes to 0.

Using only the leading term, \( \sqrt{1+t} \approx 1 + \frac{t}{2} \), greatly simplifies calculations and helps in evaluating limits.

Simplifying fractions is a key step in evaluating complex expressions. In limit problems, you often encounter expressions involving subtraction or addition of fractions. For this exercise, you have:\[ \frac{1}{t\sqrt{1+t}} - \frac{1}{t} \] The goal is to combine these into a single fraction, which involves these steps:

• Identify the common denominator.
• Subtract the numerators while keeping the common denominator.

This results in a new expression:\[ \frac{\sqrt{1+t} - 1}{t\sqrt{1+t}} \]This type of simplification helps reduce the complexity, making it easier to apply further calculus techniques like the binomial approximation. With the numerator simplified to a form that can be further approximated, this paves the way to effectively evaluate limits.

Evaluating limits is a central concept in calculus, particularly when dealing with functions that can lead to indeterminate forms like \( \frac{0}{0} \). The process involves determining the behavior of a function as a variable approaches a particular value.In the given problem, after simplification, the expression becomes:\[ \frac{1}{2\sqrt{1+t}} \]Here, as \( t \) approaches 0, \( \sqrt{1+t} \) approaches 1. Substituting this back into the expression gives:

• The limit becomes \( \frac{1}{2 \cdot 1} = \frac{1}{2} \)

Thus, by handling simplifications and approximations carefully, you determine that the limit of the initial expression is \( \frac{1}{2} \). This showcases the power of techniques like binomial approximation and effective fraction simplification in resolving such limit problems.