Question

Prove the following statements to establish the fact that $limf(x)=L$ if and only if $\underset{x\to a^{-}}{lim}f(x)=L$ and $\underset{x\to a^{+}}{lim}f(x)=L$. a. If $\underset{x\to a^{-}}{lim}f(x)=L$ and $\underset{x\to a^{+}}{lim}f(x)=L,$ then $\underset{x\to a}{lim}f(x)=L$ b. If $\underset{x\to a}{lim}f(x)=L,$ then $\underset{x\to a^{-}}{lim}f(x)=L$ and $\underset{x\to a^{+}}{lim}f(x)=L$

Short answer

Question: Prove that the limit of a function at a point exists if and only if both the left-hand limit and the right-hand limit at that point exist and are equal. Answer: To prove this statement, we have shown that if the left-hand limit and the right-hand limit both exist and are equal, then the limit at that point exists (Statement a). Conversely, we have also shown that if the limit of a function at a point exists, then both the left-hand limit and the right-hand limit exist and are equal (Statement b). These two statements together establish the fact that the limit of a function at a point exists if and only if both the left-hand limit and the right-hand limit at that point exist and are equal.

Step-by-step solution

Show that if left-hand and right-hand limits exist and are equal, then the limit exists

We are given that $\underset{x\to a^{-}}{lim}f(x)=L$ and $\underset{x\to a^{+}}{lim}f(x)=L$. We need to show that $\underset{x\to a}{lim}f(x)=L$. To prove this, let's consider any sequence $(x_n)$ converging to $a$. Since the left-hand limit exists, for any subsequence $(x_n^{\prime })$ of $(x_n)$ where $x_n^{\prime }<a$, we have $\underset{n\to \mathrm{\infty }}{lim}f(x_n^{\prime })=L$. Similarly, since the right-hand limit exists, for any subsequence $(x_n^{″})$ of $(x_n)$ where $x_n^{″}>a$, we have $\underset{n\to \mathrm{\infty }}{lim}f(x_n^{″})=L$. Since for any sequence $(x_n)$ converging to $a$, we can always find subsequences $(x_n^{\prime })$ and $(x_n^{″})$ such that they satisfy the above conditions, we have $\underset{x\to a}{lim}f(x)=L$. This proves statement a. #Statement b#

Show that if the limit exists, then the left-hand and right-hand limits exist and are equal

We are given that $\underset{x\to a}{lim}f(x)=L$. We need to show that $\underset{x\to a^{-}}{lim}f(x)=L$ and $\underset{x\to a^{+}}{lim}f(x)=L$. Let's first prove the left-hand limit: We need to show that for any sequence $(x_n)$ where $x_n<a$ and converging to $a$, $\underset{n\to \mathrm{\infty }}{lim}f(x_n)=L$. Since $\underset{x\to a}{lim}f(x)=L$, for the sequence $(x_n)$ converging to $a$, $\underset{n\to \mathrm{\infty }}{lim}f(x_n)=L$. So, $\underset{x\to a^{-}}{lim}f(x)=L$. Now let's prove the right-hand limit: We need to show that for any sequence $(x_n)$ where $x_n>a$ and converging to $a$, $\underset{n\to \mathrm{\infty }}{lim}f(x_n)=L$. Since $\underset{x\to a}{lim}f(x)=L$, for the sequence $(x_n)$ converging to $a$, $\underset{n\to \mathrm{\infty }}{lim}f(x_n)=L$. So, $\underset{x\to a^{+}}{lim}f(x)=L$. This proves statement b and completes the solution.

Key concepts

Left-hand limit

The left-hand limit is a fundamental concept in calculus used to determine the behavior of a function as it approaches a certain point from the left side. In mathematical notation, it is expressed as $\underset{x\to a^{-}}{lim}f(x)=L$, indicating that as $x$ approaches $a$ from values less than $a$, the function $f(x)$ gets closer and closer to $L$.

Understanding the left-hand limit is crucial since it helps in analyzing the continuity and differentiability of functions. If the left-hand limit at a point exists, it means

• There is a consistent, predictable pattern in the function's values when approached from the left side.
• It is possible for a function to have a left-hand limit without an overall limit existing.

One way of verifying the left-hand limit is by taking sequences that approach $a$ from the left and checking if they lead the function $f(x)$ consistently towards $L$. This aligns with the exercise where sequences and subsequences play a role in demonstrating this limit.

Right-hand limit

Just like the left-hand limit, the right-hand limit examines the behavior of a function as it approaches a point, but from the right. It is denoted mathematically by $\underset{x\to a^{+}}{lim}f(x)=L$, which means as x approaches $a$ from values greater than $a$, the function $f(x)$ approaches $L$.

The right-hand limit is essential when evaluating the continuity and overall limits of a function, particularly in pinpointing where potential discontinuities lie. To further understand right-hand limits:

• The function must have predictable behavior when approached from the right.
• Different right and left-hand limits indicate a discontinuity at $a$.

To prove a right-hand limit, sequences that approach $a$ from the right are evaluated. As shown in the exercise, combining this with the left-hand limit allows one to confirm the existence of the overall limit of a function at a point.

Sequence convergence

Sequence convergence is an integral part of understanding limits. It refers to whether a sequence approaches a specific value as it progresses infinitely. A sequence $(x_n)$ is said to converge to $a$ if, as $n$ becomes extremely large, $x_n$ gets arbitrarily close to $a$.

When we relate sequence convergence to limit theory, it allows us to establish the limits of functions by utilizing sequences. In terms of function limits:

• If a sequence of values closer and closer to $a$ leads $f(x)$ to $L$, then the limit at $x=a$ can be established.
• By examining subsequences (like those approaching from either side of $a$), we can validate left-hand and right-hand limits.

In the exercise, sequence convergence helps to demonstrate the existence and equality of left-hand and right-hand limits, therefore establishing the overall limit $\underset{x\to a}{lim}f(x)=L$. Knowing how sequences operate provides a more intuitive and practical look at the concept of limits.