Problem 77
Question
Proof Prove that $$\lim _{x \rightarrow c} f(x)=L$$ is equivalent to $$\lim _{x \rightarrow c}[f(x)-L]=0$$
Step-by-Step Solution
Verified Answer
The statement \( \lim _{x \rightarrow c} f(x)=L \) is equivalent to \( \lim _{x \rightarrow c}[f(x)-L]=0 \) because as the function f(x) gets closer to the limit L, the expression \(f(x)-L\) gets closer to zero.
1Step 1: Understanding of Limit Properties
Recall that the limit as x approaches a value is the value the function approaches. If \( \lim _{x \rightarrow c} f(x)=L \), this means that as x gets closer to c, f(x) gets closer to L.
2Step 2: Subtracting Limit L from f(x)
Consider \(f(x) - L\). As x gets infinitely close to c, \(f(x)\) gets infinitely close to L, the difference \(f(x)-L\) would then get infinitely close to 0, which means that \( \lim _{x \rightarrow c}[f(x)-L]=0 \)
3Step 3: Summary
So, \( \lim _{x \rightarrow c} f(x)=L \) is proven to be equivalent to \( \lim _{x \rightarrow c}[f(x)-L]=0 \)
Key Concepts
Limit PropertiesEquivalence of LimitsLimit Proof
Limit Properties
Limit properties form the cornerstone of understanding when studying calculus. Limits describe how a function behaves as it approaches a particular point.
For instance, let's consider the expression \( \lim_{x \rightarrow c} f(x) = L \). This states that as \( x \) gets very close to \( c \), the function \( f(x) \) approaches a value \( L \).
This concept is fundamental as it defines continuity and helps analyze function behavior even at points not explicitly included in the domain.
Some key properties of limits include:
For instance, let's consider the expression \( \lim_{x \rightarrow c} f(x) = L \). This states that as \( x \) gets very close to \( c \), the function \( f(x) \) approaches a value \( L \).
This concept is fundamental as it defines continuity and helps analyze function behavior even at points not explicitly included in the domain.
Some key properties of limits include:
- Linearity: If \( \lim_{x \rightarrow c} f(x) = L \) and \( \lim_{x \rightarrow c} g(x) = M \), then \( \lim_{x \rightarrow c} [af(x) + bg(x)] = aL + bM \) for any constants \( a \) and \( b \).
- Product Rule: \( \lim_{x \rightarrow c} [f(x) \cdot g(x)] = L \cdot M \).
- Quotient Rule: If \( M eq 0 \), \( \lim_{x \rightarrow c} \frac{f(x)}{g(x)} = \frac{L}{M} \).
Equivalence of Limits
When we talk about the equivalence of limits, we refer to the concept that if two limit expressions evaluate to the same value, they reflect the same limit behavior. This is precisely what we need to demonstrate with:
- \( \lim _{x \rightarrow c} f(x) = L \) - and \( \lim _{x \rightarrow c} [f(x) - L] = 0 \).The logic is straightforward. If \( f(x) \) approaches \( L \) as \( x \) approaches \( c \), then logically, the function \( f(x) - L \) should approach 0 when \( x \) approaches \( c \) as well. This means:
- \( \lim _{x \rightarrow c} f(x) = L \) - and \( \lim _{x \rightarrow c} [f(x) - L] = 0 \).The logic is straightforward. If \( f(x) \) approaches \( L \) as \( x \) approaches \( c \), then logically, the function \( f(x) - L \) should approach 0 when \( x \) approaches \( c \) as well. This means:
- Both expressions consider the behavior of \( f(x) \) near \( c \).
- The difference \( f(x) - L \) getting close to 0 confirms that \( f(x) \) itself gets close to \( L \).
Limit Proof
To prove the equivalence of the limits \( \lim _{x \rightarrow c} f(x) = L \) and \( \lim _{x \rightarrow c} [f(x) - L] = 0 \), we can follow a simple step-by-step logical argument.
Start by considering \( f(x) \) reaching \( L \) as \( x \) nears \( c \).
Thus, for any chosen closeness \( \epsilon > 0 \), there exists a margin \( \delta > 0 \) such that whenever \( 0 < |x - c| < \delta \), we have \( |f(x) - L| < \epsilon \).
Now, think about \( f(x) - L \). If \( |f(x) - L| < \epsilon \), it implies \( |f(x) - L - 0| < \epsilon \).
This shows that as \( x \) approaches \( c \), \( f(x) - L \) indeed approaches 0.
In this proof, every step solidifies the idea that limiting towards \( 0 \) for \( f(x) - L \) is equivalent to \( f(x) \) reaching \( L \).
Hence, the original and altered expressions are confirmed to convey the same mathematical reality.
Start by considering \( f(x) \) reaching \( L \) as \( x \) nears \( c \).
Thus, for any chosen closeness \( \epsilon > 0 \), there exists a margin \( \delta > 0 \) such that whenever \( 0 < |x - c| < \delta \), we have \( |f(x) - L| < \epsilon \).
Now, think about \( f(x) - L \). If \( |f(x) - L| < \epsilon \), it implies \( |f(x) - L - 0| < \epsilon \).
This shows that as \( x \) approaches \( c \), \( f(x) - L \) indeed approaches 0.
In this proof, every step solidifies the idea that limiting towards \( 0 \) for \( f(x) - L \) is equivalent to \( f(x) \) reaching \( L \).
Hence, the original and altered expressions are confirmed to convey the same mathematical reality.
Other exercises in this chapter
Problem 76
Graphical, Numerical, and Analytic Analysis In Exercises \(75-82\) , use a graphing utility to graph the function and estimate the limit. Use a table to reinfor
View solution Problem 77
Testing for Continuity In Exercises \(77-84\) , describe the interval(s) on which the function is continuous. $$ f(x)=\frac{x}{x^{2}+x+2} $$
View solution Problem 77
Graphical, Numerical, and Analytic Analysis In Exercises \(75-82\) , use a graphing utility to graph the function and estimate the limit. Use a table to reinfor
View solution Problem 78
Testing for Continuity In Exercises \(77-84\) , describe the interval(s) on which the function is continuous. $$ f(x)=\frac{x+1}{\sqrt{x}} $$
View solution