Problem 62 - Precalculus with Limits

Question

In Exercises 49-68, find the limit by direct substitution. $$ \lim_{x \to 8}\ \dfrac{\sqrt{x+1}}{x-4}$$

Step-by-Step Solution

Answer

\(\lim_{x \to 8}\ \dfrac{\sqrt{x+1}}{x-4} = \dfrac{3}{4}\)

Step 1: Check for Indeterminate Forms

Before directly substituting the value of \(x\) in the expression, it's crucial to check if the function evaluates into an indeterminate form. By substituting \(x = 8\) in the denominator \(x - 4\), we get \(8 - 4\) which is equal to 4, not zero. Therefore, we don't have an indeterminate form, and we can proceed with direct substitution.

Step 2: Direct Substitution

Now, substitute \(x = 8\) in the expression \(\dfrac{\sqrt{x+1}}{x-4}\). Thus, the expression becomes \(\dfrac{\sqrt{8+1}}{8-4}\)

Step 3: Simplify the expression

Simplify the expression to get the limit. Therefore, the expression \(\dfrac{\sqrt{8+1}}{8-4}\) simplifies to \(\dfrac{\sqrt{9}}{4} = \dfrac{3}{4}\)

Key Concepts

Direct Substitution

Direct substitution is a straightforward approach to finding limits. When you substitute a specific value of \( x \) into a function, you are checking to see if you can directly determine the limit without additional manipulation. In this context, substituting helps us evaluate expressions close to the point we are interested in, particularly as \( x \) approaches the number specified in the limit problem.

Before applying direct substitution, it's vital to check if the function allows this method. Plugging the value directly into expression should not result in an undefined or indeterminate form. If it does not result in such issues, it is a quick and effective way to find limits.

For example, in our exercise, we tested by substituting \( x = 8 \) into the denominator \( x - 4 \). Since the outcome was a non-zero value (4), we could proceed comfortably with direct substitution.

Indeterminate Forms

Indeterminate forms are mathematical expressions that do not readily provide a clear limit, usually arising when directly substituting values into functions. Common indeterminate forms include \( \frac{0}{0} \), \( \infty - \infty \), and \( \infty/\infty \), among others.

These forms can mislead you in determining the true behavior of the function as \( x \) approaches a specific value. Therefore, recognizing indeterminate forms is essential to decide if direct substitution can be applied, or if a different strategy needs to be implemented to resolve the limit.

In our example, by checking \( x - 4 \) when \( x = 8 \), we confirmed that it evaluates to 4. This result reassures us that the form is determinate, negating any need for more advanced limit-solving techniques and allowing us to proceed smoothly with direct substitution.

Simplification

Simplification is the process of reducing an expression to its simplest form. This process aids in making calculations easier and reveals more about the function's behavior.

Once the substitution has been applied, simplifying the resulting expression helps in confirming the limit value. For polynomial expressions, it might involve factoring or canceling terms, whereas for fractions or roots, similar simplification methods are used.

In our example, after substituting \( x = 8 \), the expression \( \frac{\sqrt{8+1}}{8-4} \) becomes \( \frac{\sqrt{9}}{4} \). Simplifying further to \( \frac{3}{4} \) makes the limit clear and ensures that all unnecessary complexity is removed from the expression.