Question

Use the following definitions. Assume fexists for all \(x\) near a with \(x>\) a. We say the limit of \(f(x)\) as \(x\) approaches a from the right of a is \(L\) and write \(\lim _{x \rightarrow a^{+}} f(x)=L,\) if for any \(\varepsilon>0\) there exists \(\delta>0\) such that $$|f(x)-L|<\varepsilon \quad \text { whenever } \quad 0< x-a<\delta$$ Assume fexists for all \(x\) near a with \(x < \) a. We say the limit of \(f(x)\) as \(x\) approaches a from the left of a is \(L\) and write \(\lim _{x \rightarrow a^{-}} f(x)=L,\) if for any \(\varepsilon > 0 \) there exists \(\delta > 0\) such that $$|f(x)-L| < \varepsilon \quad \text { whenever } \quad 0< a-x <\delta$$ Prove the following limits for $$f(x)=\left\\{\begin{array}{ll} 3 x-4 & \text { if } x<0 \\ 2 x-4 & \text { if } x \geq 0 \end{array}\right.$$ a. \(\lim _{x \rightarrow 0^{+}} f(x)=-4\) b. \(\lim _{x \rightarrow 0^{-}} f(x)=-4\) c. \(\lim _{x \rightarrow 0} f(x)=-4\)

Short answer

Short Answer: Given a function $f(x)$ defined as: $$f(x)=\left\\{\begin{array}{ll} 3 x-4 & \text { if } x<0 \\\ 2 x-4 & \text { if } x \geq 0 \end{array}\right.$$ We proved the right-hand limit, left-hand limit, and overall limit when $x \rightarrow 0$ to be $-4$. We used the ε-δ definition of limits to show that the right-hand limit exists when $x \geq 0$ for the function $2x-4$, selecting $\delta=\varepsilon/2$. Similarly, we showed that the left-hand limit exists when $x < 0$ for the function $3x-4$, selecting $\delta=\varepsilon/3$. Since both the right-hand and left-hand limits exist and are equal to $-4$, the overall limit when $x \rightarrow 0$ is $-4$.

Step-by-step solution

1. Function for x ≥ 0

Since \(x\geq0\), we will only consider the part of the function which says \(f(x)=2x-4\).

2. ε-δ Definition

According to the ε-δ definition of right-hand limit, \(\lim _{x \rightarrow 0^{+}} f(x)=-4\) if for any \(\varepsilon>0\) there exists a \(\delta>0\) such that whenever \(0 < x-0 < \delta\), we have \(|f(x)-(-4)|<\varepsilon\).

3. Substitute f(x)

Now, substitute f(x) with \(2x-4\), and show that \(|2x-4-(-4)|<\varepsilon\) whenever \(0 < x < \delta\).

4. Simplify and Solve for δ

Simplify the inequality as follows: $$|2x-4+4| = |2x| < \varepsilon$$ To find a suitable \(\delta\), set \(\delta=\varepsilon/2\). So, whenever \(0 < x < \delta=\frac{\varepsilon}{2}\), we have \(|2x|<\varepsilon\). Thus, the limit is proven: \(\lim _{x \rightarrow 0^{+}} f(x)=-4\). b. Prove that \(\lim _{x \rightarrow 0^{-}} f(x)=-4\)

1. Function for x < 0

Since \(x<0\), we will only consider the part of the function which says \(f(x)=3x-4\).

2. ε-δ Definition

According to the ε-δ definition of left-hand limit, \(\lim _{x \rightarrow 0^{-}} f(x)=-4\) if for any \(\varepsilon>0\) there exists a \(\delta>0\) such that whenever \(0 < 0-x < \delta\), we have \(|f(x)-(-4)|<\varepsilon\).

3. Substitute f(x)

Now, substitute f(x) with \(3x-4\), and show that \(|3x-4-(-4)|<\varepsilon\) whenever \(0 < -x < \delta\).

4. Simplify and Solve for δ

Simplify the inequality as follows: $$|3x-4+4| = |3x| < \varepsilon$$ To find a suitable \(\delta\), set \(\delta=\varepsilon/3\). So, whenever \(0 < -x < \delta=\frac{\varepsilon}{3}\), we have \(|3x|<\varepsilon\). Thus, the limit is proven: \(\lim _{x \rightarrow 0^{-}} f(x)=-4\). c. Prove that \(\lim _{x \rightarrow 0} f(x)=-4\)

1. Right-hand and Left-hand Limits

The right-hand limit for \(x \rightarrow 0\) was proven as \(-4\) in part a, and the left-hand limit for \(x \rightarrow 0\) was proven as \(-4\) in part b.

2. Equality of Right-hand and Left-hand limits

According to the definition of a function's limit, if the right-hand limit and the left-hand limit are equal, then the limit exists. In this case, both limits are equal to \(-4\). Thus, the limit exists: \(\lim _{x \rightarrow 0} f(x)=-4\).

Key concepts

Epsilon-Delta Definition

The \(\varepsilon-\delta\) definition forms the backbone of understanding limits in calculus. It's all about creating a precise framework to define how a function behaves as it approaches a particular point. \(\epsilon\) (epsilon) represents any small positive number we choose, indicating how close the function's value, \(f(x)\), is expected to be to the limit \(L\). On the other hand, \(\delta\) (delta) signifies a small positive number that determines how close the input values, \(x\), need to be to the target \(a\). This definition emphasizes that for every \(\epsilon>0\), there exists a \(\delta>0\) such that if \(0<|x-a|<\delta\), then \(|f(x)-L|<\epsilon\). In simpler terms, if we zoom into the point where \(x\) is close to \(a\), then \(f(x)\) should be close to \(L\) within our chosen proximity defined by \(\epsilon\). This criterion gets applied in proving both right-hand and left-hand limits.

Right-Hand Limit

The right-hand limit is concerned with how a function behaves as the variable \(x\) approaches a specific point \(a\) from the right-hand side, meaning values of \(x\) that are greater than \(a\). For instance, if \(f(x)\) approaches a limit \(L\) as \(x\) comes closer to \(a\) from the right (\(x>a\),), then this is described as \(\lim_{x \to a^{+}} f(x) = L\). The process of proving a right-hand limit requires showing that for every \(\epsilon>0\), you can find a \(\delta>0\) such that if \(0

Left-Hand Limit

Analogous to the right-hand limit, the left-hand limit examines the behavior of \(f(x)\) as \(x\) approaches a point \(a\) from the left, where \(x0\), a \(\delta>0\) must be found such that if \(0

Left-Hand Limit

Analogous to the right-hand limit, the left-hand limit examines the behavior of \(f(x)\) as \(x\) approaches a point \(a\) from the left, where \(x0\), a \(\delta>0\) must be found such that if \(0