91Ó°ÊÓ

In calculus, when evaluating limits, you might encounter expressions that result in forms such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) when you try direct substitution. These are known as indeterminate forms. Indeterminate means that the form doesn’t directly tell us what the limit is, if the limit exists at all. This happens because operations like division by zero or infinity don't yield a straightforward result. Instead, we need to further manipulate the expression to find the limit. In the exercise, \( \lim _{h \rightarrow 0} \frac{\sqrt{9+h}-3}{h} \) initially gives a \( \frac{0}{0} \) form since both the numerator \( \sqrt{9+0}-3 \) and the denominator \( h \) approach zero. Therefore, the expression needs to be simplified before the limit can be properly evaluated.

Rationalization is a technique often used to simplify expressions with square roots in the numerator or denominator. By multiplying an expression by a conjugate, you can eliminate the square root terms. The conjugate of an expression \( a + b \) is \( a - b \), and vice versa. This works because the product of both forms a difference of squares, which simplifies to \( a^2 - b^2 \). In the context of the given exercise, rationalizing the numerator \( \sqrt{9+h} - 3 \) involves multiplying both the numerator and the denominator by its conjugate \( \sqrt{9+h} + 3 \). This results in a simplified expression that avoids the indeterminate form, making it possible to evaluate the limit.

The difference of squares is a fundamental algebraic identity: \((a-b)(a+b) = a^2 - b^2\). It’s particularly useful in calculus when dealing with limits that involve square roots or other polynomial expressions. When you apply this technique, it allows you to remove the square root by transforming the expression into a simpler quadratic form. In our exercise example, after rationalizing the numerator using \( \sqrt{9+h} + 3 \), the difference of squares simplifies \((\sqrt{9+h})^2 - 3^2\) which equals \((9+h) - 9 = h\). Simplification through this technique not only removes the square root but also enables easy cancellation with the denominator, simplifying the expression further.

After simplifying the expression, the final step is to evaluate the limit. With the indeterminate form resolved, we are left with an expression that allows substitution. Cancel out the common \( h \) from both the numerator and the denominator:

• Expression: \( \lim _{h \rightarrow 0} \frac{h}{h(\sqrt{9+h}+3)} \)
• Cancelled: \( \lim _{h \rightarrow 0} \frac{1}{\sqrt{9+h}+3} \)

Now substitute \( h = 0 \) into \( \frac{1}{\sqrt{9+0}+3} \), yielding \( \frac{1}{6} \). Evaluating the limit means confirming that this number is the value the function approaches as \( h \) gets arbitrarily close to 0. Through careful manipulation and simplification, we find that this limit indeed exists and equals \( \frac{1}{6} \).