Problem 28
Question
, find the limits if \(\lim _{x \rightarrow a} f(x)=3\) and \(\lim _{x \rightarrow a} g(x)=-1(\) see Example 4) \(.\) $$ \lim _{x \rightarrow a}[f(x)-3]^{4} $$
Step-by-Step Solution
Verified Answer
The limit is 0.
1Step 1: Identify Given Limits
We are given that \( \lim_{x \to a} f(x) = 3 \). This means as \( x \) approaches \( a \), the function \( f(x) \) approaches the value 3.
2Step 2: Recognize Expression Structure
The limit we need to find is \( \lim_{x \to a}[f(x) - 3]^{4} \). Alternately, this can be expressed as raising the term \( [f(x)-3] \) to the fourth power.
3Step 3: Substitute the Limit into the Expression
Since \( \lim_{x \to a} f(x) = 3 \), the expression \( f(x) - 3 \) approaches 0 as \( x \to a \). Therefore, \( \lim_{x \to a} (f(x) - 3) = 0 \).
4Step 4: Apply Limits to Power Rule
Raised to the fourth power, \( [f(x) - 3]^{4} \) becomes 0 raised to the fourth power as \( x \) approaches \( a \). This follows the power rule: \( \lim_{x \to a} (u(x))^{n} = (\lim_{x \to a} u(x))^{n} \) if the limit \( u(x) \) exists.
5Step 5: Calculate the Final Limit
Applying the power rule to our expression, we have \( \lim_{x \to a} [f(x)-3]^{4} = (\lim_{x \to a} [f(x)-3])^{4} = 0^{4} = 0 \).
Key Concepts
Power rule in calculusContinuity in calculusEvaluating limits
Power rule in calculus
The power rule in calculus provides a straightforward way to differentiate functions that are exponentials of a term. However, it is also key in the context of evaluating limits of expressions that involve powers. The rule states that, if you have a function raised to an exponent, you can evaluate its limit by firstly finding the limit of the base and then raising that result to the given power. Formally, this is written as:\[ \lim_{x \to a} (u(x))^{n} = \left( \lim_{x \to a} u(x) \right)^{n} \]where the limit \( u(x) \) exists.
- To apply the power rule, ensure the base function’s limit is well-defined and finite. Otherwise, the power rule may not apply.
- This rule is useful in simplifying expressions and computing limits that would otherwise seem complex without it.
- In our exercise, the rule allowed us to simplify \( \lim_{x \to a} [f(x)-3]^{4} \) to \( 0^4 = 0 \), since the limit of \( f(x)-3 \) was 0.
Continuity in calculus
Continuity is a fundamental property of functions in calculus that ensures a function behaves predictably around certain points. A function is continuous at a point if the following conditions are simultaneously met:
- The function is defined at that specific point.
- The limit of the function as it approaches the point is defined.
- The value of the function at that point is equal to the limit of the function as it approaches the point.
Evaluating limits
Evaluating limits involves finding the value a function approaches as the input approaches a specific point. This process can seem daunting, but there are several strategies and rules, including the power rule, that simplify this task immensely.
- Direct Substitution: Apply direct substitution if the function is continuous at the point of interest. It often provides the limit directly.
- Simplifying Expressions: Factor or use algebraic identities to simplify the expression before evaluating the limit.
- L'Hôpital's Rule: Use when you encounter indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).
Other exercises in this chapter
Problem 28
$$ \begin{array}{l} \text { Prove that if }|f(x)|
View solution Problem 28
, use a calculator to find the indicated limit. Use a graphing calculator to plot the function near the limit point. $$ \lim _{u \rightarrow \pi / 2} \frac{2-2
View solution Problem 29
What points, if any, are the functions discontinuous? $$ g(u)=\frac{u^{2}+|u-1|}{\sqrt[3]{u+1}} $$
View solution Problem 29
How long does it take money to double in value for the specified interest rate? (a) \(6 \%\) compounded monthly (b) \(6 \%\) compounded continuously
View solution