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Evaluate the indicated limits by direct evaluation as in Examples \(10-14 .\) Change the form of the function where necessary. $$\lim _{x \rightarrow 4} \sqrt{x^{2}-7}$$

Short Answer

Expert verified
The limit is 3.

Step by step solution

01

Check for Direct Substitution

First, try to evaluate the limit by direct substitution. For this limit, substitute \( x = 4 \) directly into the function: \( \sqrt{x^2 - 7} \). This gives us \( \sqrt{4^2 - 7} \).
02

Simplify the Expression

Now, simplify the expression inside the square root: \( 4^2 = 16 \), so the expression becomes \( \sqrt{16 - 7} \).
03

Perform the Calculation

Calculate the simplified expression: \( 16 - 7 = 9 \). Thus, the expression becomes \( \sqrt{9} \).
04

Find the Square Root

Finally, find the square root of 9, which is \( 3 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Direct Substitution
Calculating limits can sometimes be as simple as directly substituting the value into the function. This approach is known as **direct substitution**. It involves replacing the variable with the number it's approaching. For many functions, especially polynomials and rational functions, this approach works well provided there's no division by zero or undefined terms.
When given a function like \( \lim_{x \rightarrow 4} \sqrt{x^2 - 7} \), we attempt to substitute \( x = 4 \) directly into the equation. If the resulting expression is defined and finite, then the limit can be evaluated using this substitution.
  • Direct substitution is quick and efficient.
  • Always check if the substituted expression is defined.
  • If direct substitution results in an indeterminacy like \( \frac{0}{0} \), you'll need to apply other evaluative techniques.
Limit Evaluation
When direct substitution doesn't immediately provide an answer, it's essential to carefully evaluate the limit. **Limit evaluation** refers to the process of determining the value that a function approaches as the input approaches a certain number. This can be done by simplifying the function until it is easier to substitute, exploring alternative algebraic techniques, or recognizing existing limits.
In our example, \( \lim_{x \rightarrow 4} \sqrt{x^2 - 7} \), the function was already in a form where direct substitution sufficed. However, if it wasn't feasible due to an undefined expression, exploring factorization, rationalization, or using limit laws would be necessary.
  • Simplifying expressions is often the key to finding limits.
  • Always aim to evaluate step-by-step, ensuring the expression remains valid.
  • Understanding algebraic manipulations enhances limit evaluation skills.
Square Root Calculations
Evaluating expressions with square roots requires a careful approach, especially within the context of limits. **Square root calculations** necessitate determining the value of square roots accurately, keeping in mind that a square root of a positive number has both a principal (positive) and a negative value. However, in most calculus problems, we focus on the principal root.
In our exercise \( \lim_{x \rightarrow 4} \sqrt{x^2 - 7} \), once we've simplified the expression to \( \sqrt{9} \), we calculate the square root. Since \( 9 \) is a perfect square, its square root is simply \( 3 \). This step concludes our limit evaluation.
  • Accurate square root evaluation is pivotal in solving limits involving radicals.
  • Identify perfect squares to simplify calculations.
  • Remember to focus on the principal root unless otherwise specified.

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Most popular questions from this chapter

Solve the given problems involving limits. Velocity can be found by dividing the displacement \(s\) of an object by the elapsed time \(t\) in moving through the displacement. In a certain experiment, the following values were measured for the displacements and elapsed times for the motion of an object. Determine the limiting value of the velocity. $$\begin{array}{l|c|c|c|c|c}s(\mathrm{cm}) & 0.480000 & 0.280000 & 0.029800 & 0.0029980 & 0.00029998 \\ \hline t(\mathrm{s}) & 0.200000 & 0.100000 & 0.010000 & 0.0010000 & 0.00010000\end{array}$$

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