Problem 28 Find each limit, if it exists: ... [FREE SOLUTION]

Chapter 2: Problem 28

Find each limit, if it exists: |a) $\lim _{x \rightarrow a^{-}} f(x)$ (b) $\lim _{x \rightarrow a^{+}} f(x)$ (c) $\lim _{x \rightarrow a} f(x)$ $$ f(x)=\sqrt{5-2 x}-x^{2}: \quad a=\frac{5}{2} $$

Short Answer

Expert verified

Left-hand limit is $-\frac{25}{4}$, right-hand and two-sided limits do not exist.

Step by step solution

Understand the problem

We need to evaluate the limits of the function $f(x) = \sqrt{5-2x}-x^2$ as $x$ approaches $\frac{5}{2}$ from the left, right, and directly. These are one-sided limits and a two-sided limit at the point where $a = \frac{5}{2}$.

Determine the left-hand limit

To find $\lim_{x \to \frac{5}{2}^{-}} f(x)$, analyze the function values as $x$ approaches $\frac{5}{2}$ from the left side. The square root function $\sqrt{5-2x}$ will approach zero as $x$ approaches $\frac{5}{2}$. Also, $(-x^2)$ will become increasingly negative, going towards $-\left(\frac{5}{2}\right)^2 = -\frac{25}{4}$. Therefore, the $f(x)$ will approach $-\frac{25}{4}$.

Determine the right-hand limit

For $\lim_{x \to \frac{5}{2}^{+}} f(x)$, consider values of $x$ slightly greater than $\frac{5}{2}$. For $x > \frac{5}{2}$, the expression inside the square root $5-2x$ becomes negative, meaning $\sqrt{5-2x}$ is not real for real $x$. As a result, the function $f(x)$ is not defined for real numbers in this case when approaching from the right, so the limit does not exist.

Evaluate the two-sided limit

The two-sided limit $\lim_{x \to \frac{5}{2}} f(x)$ only exists if both the left and right-hand limits exist and are equal. Since the right-hand limit does not exist, the two-sided limit does not exist either.

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

One-Sided Limits

In calculus, a one-sided limit refers to the behavior of a function as the input approaches a specific value from only one side. This involves either approaching from the left or the right.

Left-hand limit: This is denoted as $\lim_{x \to a^-} f(x)$. Here, "$a^-$" indicates approaching $a$ from values less than $a$.
Right-hand limit: Denoted as $\lim_{x \to a^+} f(x)$, where "$a^+$" indicates that $x$ is approaching $a$ from values greater than $a$.

For our function, $f(x) = \sqrt{5 - 2x} - x^2$, as $x$ nears $\frac{5}{2}$ from the left, the function tends toward $-\frac{25}{4}$. However, if we approach from the right, the function isn't defined because the square root becomes imaginary, leading to no limit.

Two-Sided Limits

A two-sided limit considers approaching a point from both directions, left and right. For a two-sided limit $\lim_{x \to a} f(x)$ to exist, both the left-hand limit $\lim_{x \to a^-} f(x)$ and the right-hand limit $\lim_{x \to a^+} f(x)$ must exist and be the same.
In simple terms, for the function to have a clear limit at any given point, approaching from either side must yield the same value.
In our case of $f(x) = \sqrt{5 - 2x} - x^2$, as we saw:

The left-hand limit exists and equals $-\frac{25}{4}$.
The right-hand limit does not exist due to undefined values stemming from negative square roots.

Since the limits differ, the overall or two-sided limit at $x = \frac{5}{2}$ cannot exist.

Function Continuity

A function is continuous at a point if it has no jumps, breaks, or holes at that point. Mathematically, a function $f(x)$ is continuous at $x = a$ if three conditions hold:

$f(a)$ is defined.
The limit $\lim_{x \to a} f(x)$ exists.
The value of the function equals the limit at that point, i.e., $f(a) = \lim_{x \to a} f(x)$.

If any condition fails, the function is not continuous. In our function $f(x) = \sqrt{5 - 2x} - x^2$, there is a problem at $x = \frac{5}{2}$ because the two-sided limit doesn't exist. The inconsistency between sides causes the function's value to be undefined right at the boundary, indicating a discontinuity.

Square Root Function

The square root function, represented by $\sqrt{x}$, is a fundamental part of algebra involving the extraction of roots. It's important in calculus to understand its domain, which is crucial for solving limit problems.
Generally, $\sqrt{x}$ is defined for $x \geq 0$. This impacts expressions like $\sqrt{5 - 2x}$. The function $f(x) = \sqrt{5 - 2x} - x^2$ is directly influenced by the nature of square roots.

For $5 - 2x \geq 0$, $f(x)$ remains real, typically yielding real number outputs.
For $5 - 2x < 0$, the square root is undefined in the real number sense, causing the functional limit to cease existing on the right of $x = \frac{5}{2}$.

Due to such intricacies, examining the behavior of $\sqrt{5 - 2x}$ directly affects our understanding of function limits and continuity.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

91影视

Find each limit, if it exists: |a) \(\lim _{x \rightarrow a^{-}} f(x)\) (b) \(\lim _{x \rightarrow a^{+}} f(x)\) (c) \(\lim _{x \rightarrow a} f(x)\) $$ f(x)=\sqrt{5-2 x}-x^{2}: \quad a=\frac{5}{2} $$

Short Answer

Step by step solution

Understand the problem

Determine the left-hand limit

Determine the right-hand limit

Evaluate the two-sided limit

Key Concepts

One-Sided Limits

Two-Sided Limits

Function Continuity

Square Root Function

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Discrete Mathematics

Geometry

Mechanics Maths

Calculus

Statistics

Probability and Statistics

Study anywhere. Anytime. Across all devices.