Problem 20
Question
Find the limits in Exercises \(11-22\) $$\lim _{z \rightarrow 4} \sqrt{z^{2}-10}$$
Step-by-Step Solution
Verified Answer
The limit is \(\sqrt{6}\).
1Step 1: Evaluate function at the limit point
Substitute the value of the limit point directly into the function. The function is given by \(f(z) = \sqrt{z^2 - 10}\). For \(z \to 4\), we substitute \(z = 4\) into the function: \(f(4) = \sqrt{4^2 - 10}\).
2Step 2: Simplify the expression
Simplify the expression inside the square root. First, calculate \(4^2 = 16\). Therefore, the expression becomes \(\sqrt{16 - 10}\).
3Step 3: Perform subtraction
Subtract \(10\) from \(16\), resulting in \(6\). Thus, our expression becomes \(\sqrt{6}\).
4Step 4: Determine the limit
The limit as \(z\) approaches \(4\) of \(\sqrt{z^2 - 10}\) is the value of the simplified expression: \(\sqrt{6}\). Since \(\sqrt{6}\) is a constant value, the limit is \(\sqrt{6}\).
Key Concepts
Continuous FunctionsEvaluation of LimitsLimit Laws
Continuous Functions
In calculus, a continuous function is one where you can draw it without lifting your pen off the paper. This is because at every point in its domain, the function doesn't have any holes or jumps.
For a function like \(f(z) = \sqrt{z^2 - 10}\), determining continuity involves ensuring that the expression inside the square root is non-negative in the relevant domain. At \(z = 4\), the expression is \(16 - 10 = 6\), which is positive. Therefore, our function is continuous at this point.
This property of continuity often allows us to directly substitute the value of \(z\) to find the limit, making computations simpler and straightforward.
For a function like \(f(z) = \sqrt{z^2 - 10}\), determining continuity involves ensuring that the expression inside the square root is non-negative in the relevant domain. At \(z = 4\), the expression is \(16 - 10 = 6\), which is positive. Therefore, our function is continuous at this point.
This property of continuity often allows us to directly substitute the value of \(z\) to find the limit, making computations simpler and straightforward.
Evaluation of Limits
Evaluating limits is a fundamental concept in calculus, helping us understand the behavior of a function as it approaches a certain point.
The evaluation process typically involves directly substituting the approaching value into the function, provided the function is continuous at the point, and simplifying the resulting expression.
For example, in \(\lim _{z \rightarrow 4} \sqrt{z^2-10}\), we replace \(z\) with \(4\), making the expression \(\sqrt{4^2 - 10}\).
The evaluation process typically involves directly substituting the approaching value into the function, provided the function is continuous at the point, and simplifying the resulting expression.
For example, in \(\lim _{z \rightarrow 4} \sqrt{z^2-10}\), we replace \(z\) with \(4\), making the expression \(\sqrt{4^2 - 10}\).
- Calculate \(4^2 = 16\).
- Subtract to get \(16 - 10 = 6\).
- The expression simplifies to \(\sqrt{6}\).
Limit Laws
Limit laws provide us with a set of rules to make the process of finding a limit much easier. These laws apply to multiple operations involving limits, such as addition, subtraction, multiplication, and division.
One helpful law is that the limit of a composition can be found by taking the limit of the inner function first, before applying the outer functions. For example, if combining square roots and polynomials, finding limits becomes more straightforward.
Another critical law is the Direct Substitution Property, which states that if a function is continuous at a point, the limit at that point can be found by directly substituting the point's value into the function.
One helpful law is that the limit of a composition can be found by taking the limit of the inner function first, before applying the outer functions. For example, if combining square roots and polynomials, finding limits becomes more straightforward.
Another critical law is the Direct Substitution Property, which states that if a function is continuous at a point, the limit at that point can be found by directly substituting the point's value into the function.
- This allows effective limit evaluation without additional algebraic manipulation when the function is smooth and without discontinuities.
Other exercises in this chapter
Problem 20
In Exercises \(13-22,\) find the limit of each rational function (a) as \(x \rightarrow \infty\) and \((b)\) as \(x \rightarrow-\infty\) . $$g(x)=\frac{x^{3}+7
View solution Problem 20
Each of Exercises \(15-30\) gives a function \(f(x)\) and numbers \(L, c,\) and \(\epsilon>0 .\) In each case, find an open interval about \(c\) on which the in
View solution Problem 21
At what points are the functions in Exercises 13-30 continuous? $$y=\csc 2 x$$
View solution Problem 21
In Exercises \(13-22,\) find the limit of each rational function (a) as \(x \rightarrow \infty\) and \((b)\) as \(x \rightarrow-\infty\) . $$f(x)=\frac{3 x^{7}+
View solution