Question

$$\text { Suppose } \lim _{x \rightarrow a} f(x)=L \text { and } \lim _{x \rightarrow a} g(x)=M$$ .Prove that \(\lim _{x \rightarrow a}(f(x)-g(x))=L-M\)

Short answer

Question: Prove that if $$\lim _{x \rightarrow a} f(x)=L$$ and $$\lim _{x \rightarrow a} g(x)=M$$, then $$\lim _{x \rightarrow a}(f(x)-g(x))=L-M$$.

Step-by-step solution

State the limit definitions for f(x) and g(x)

Given that $$\lim _{x \rightarrow a} f(x)=L$$ and $$\lim _{x \rightarrow a} g(x)=M$$, we know that for any ε > 0, there exists some $$δ_1$$ and $$δ_2$$ such that if |x - a| < $$δ_1$$, then |f(x) - L| < $$\frac{\epsilon}{2}$$, and if |x - a| < $$δ_2$$, then |g(x) – M| < $$\frac{\epsilon}{2}$$.

Establish a common δ value

Let's take δ = min{$$δ_1$$, $$δ_2$$}. Now if |x - a| < δ, we know that both |f(x) - L| < $$\frac{\epsilon}{2}$$ and |g(x) – M| < $$\frac{\epsilon}{2}$$ hold.

Apply the Triangle Inequality

Our goal is to show that |(f(x) - g(x)) - (L - M)| < ε. Let's apply the Triangle Inequality to this expression: $$|(f(x) - g(x)) - (L - M)| = |(f(x) - L) - (g(x) - M)| \leq |f(x) - L| + |g(x) - M|$$

Substitute ε values and obtain the result

From step 2, we know that |f(x) - L| < $$\frac{\epsilon}{2}$$ and |g(x) - M| < $$\frac{\epsilon}{2}$$ when |x - a| < δ. Thus, we can substitute these values into the inequality we derived in step 3: $$|(f(x) - g(x)) - (L - M)| \leq \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$

Conclusion

Since we found that |(f(x) - g(x)) - (L - M)| < ε when |x - a| < δ, we proved that $$\lim _{x \rightarrow a}(f(x)-g(x))=L-M$$.

Key concepts

Limit Laws

In calculus, limit laws provide us with tools to evaluate limits systematically. They allow us to break down complex expressions into simpler parts. For example, if you know the limits of two functions individually, you can find the limit of their sum, difference, product, or quotient using these laws.

In the problem, since \(\lim _{x \rightarrow a} f(x)=L\) and \(\lim _{x \rightarrow a} g(x)=M\), the limit laws state that the limit of their difference will be \(L-M\).

• The Sum Law: \(\lim _{x \rightarrow a} (f(x) + g(x)) = L + M\).
• The Difference Law: \(\lim _{x \rightarrow a} (f(x) - g(x)) = L - M\).
• The Product Law: \(\lim _{x \rightarrow a} (f(x) \cdot g(x)) = L \cdot M\).
• The Quotient Law: \(\lim _{x \rightarrow a} \frac{f(x)}{g(x)} = \frac{L}{M}\), assuming \ M eq 0\.

Limit laws are foundational because they allow us to work with limits in a straightforward way without having to recompute everything from scratch.

Epsilon-Delta Definition

The epsilon-delta definition is a formal way to understand limits. It precisely explains what it means for a function to approach a limit as it gets close to a certain point.

For \(\lim _{x \rightarrow a} f(x) = L\), the definition states:

• For every \( \epsilon > 0\), there exists a \( \delta > 0\) such that \( |x - a| < \delta \) implies \( |f(x) - L| < \epsilon\).

In the problem, we use delta values \(\delta_1\) and \(\delta_2\) to establish small intervals where \(f(x)\) and \(g(x)\) remain close to their respective limits, \(L\) and \(M\). By taking \(\delta = \min(\delta_1, \delta_2)\), we ensure both inequalities hold simultaneously.

This step guarantees that when \(x\) is within this interval from \(a\), both \(f(x)\) and \(g(x)\) closely approach \(L\) and \(M\), respectively. This precision allows us to confirm their difference approaches \(L-M\).

Continuity

Continuity is about understanding how a function behaves at a certain point. A function is continuous at a point if the function’s limit as it approaches that point is equal to the function’s value at the point.

In simple terms:

• A continuous function has no breaks, jumps, or holes.
• For \(f(x)\) to be continuous at \(x = a\), the following must be true: \(\lim _{x \rightarrow a} f(x) = f(a)\).

In the given exercise, examining the continuity ensures that \(f(x)\) and \(g(x)\) don't have any interruptions. This is key in applying limit laws effectively.

When functions are continuous, as in the given problem, we can confidently apply their limits in various operations, guaranteeing results like \(\lim _{x \rightarrow a} (f(x) - g(x)) = L - M\).

Triangle Inequality

The Triangle Inequality is a crucial tool in proving limits with precision. It expresses a rule about the absolute values of sums or differences.

The Triangle Inequality states:

• For any real numbers \(a\) and \(b\), \(|a + b| \leq |a| + |b|\).

In this exercise, it was applied to the expression \( |(f(x) - g(x)) - (L - M)| \). The inequality becomes:

• \( |(f(x) - L) - (g(x) - M)| \leq |f(x) - L| + |g(x) - M|\).

This allows us to combine the two conditions: \( |f(x) - L| < \frac{\epsilon}{2} \) and \( |g(x) - M| < \frac{\epsilon}{2} \), thereby concluding that \( |(f(x) - g(x)) - (L - M)| < \epsilon\).

By applying the Triangle Inequality, we show that our combined limits respect the original limits, concluding the proof for \(\lim _{x \rightarrow a} (f(x) - g(x)) = L - M\).