Problem 26

Question

\text { Prove that } \lim _{x \rightarrow 0^{+}} \sqrt{x}=0

Step-by-Step Solution

Verified

Answer

The limit $ \lim_{x \to 0^{+}} \sqrt{x} = 0 $ is proven by choosing $ \delta = \epsilon^2 $.

1Step 1: Understand the Limit Definition

We need to prove that $ \lim_{x \to 0^{+}} \sqrt{x} = 0 $. This means for every $ \epsilon > 0 $, there exists a $ \delta > 0 $ such that if $ 0 < x < \delta $, then $ \sqrt{x} < \epsilon $.

2Step 2: Express the Problem in Terms of $\epsilon$

Start by expressing $ \sqrt{x} < \epsilon $. Squaring both sides, we get $ x < \epsilon^2 $. This expression will help us find $ \delta $ in terms of $ \epsilon $.

3Step 3: Choose $ \delta $

To satisfy the limit definition, choose $ \delta = \epsilon^2 $. This choice ensures that for every $ \epsilon > 0 $, if $ 0 < x < \delta $, then $ x < \epsilon^2 $, guaranteeing $ \sqrt{x} < \epsilon $.

4Step 4: Verify the Result

Given this choice of $ \delta $, whenever $ 0 < x < \delta $, the condition $ \sqrt{x} < \epsilon $ holds true. Thus, the limit $ \lim_{x \to 0^{+}} \sqrt{x} = 0 $ is proven.

Key Concepts

Epsilon-Delta DefinitionSquare Root FunctionLimit Approaching Zero

Epsilon-Delta Definition

The epsilon-delta definition is a fundamental concept used in calculus to rigorously establish the limit of a function. In simple terms, it's a precise way to say that a function approaches a specific value as the input approaches some point. Let's break it down:

Epsilon ($\epsilon$): Represents how close you want the function's output to be to the limit value.
Delta ($\delta$): Represents how close the inputs need to be to the target point where we are taking the limit.

The basic goal is to show that for any small positive number $\epsilon$, there exists a matching small positive number $\delta$ such that if the distance between the function’s input and the limit point is less than $\delta$, then the distance between the function’s output and the limit value is less than $\epsilon$.
In the exercise, our limit is $\lim_{x \to 0^{+}} \sqrt{x} = 0$. For every $\epsilon > 0$, we must find a $\delta > 0$ ensuring if $0 < x < \delta$, then $\sqrt{x} < \epsilon$. By choosing $\delta = \epsilon^2$, this requirement is satisfied, proving the limit.

Square Root Function

The square root function is one of the most essential functions in mathematics, often denoted as $\sqrt{x}$. It maps non-negative numbers to their square roots. Here are some key features:

It's only defined for non-negative inputs, meaning $x \geq 0$.
The output of a square root function is always non-negative, making it a real-valued function.
It grows slower compared to a linear function because $\sqrt{x}$ becomes smaller as $x$ gets closer to zero.

In the context of limits, the behavior near zero is crucial. When $x \to 0^{+}$, $\sqrt{x}$ approaches zero, but does not become negative. This gradual decrease of $\sqrt{x}$ as $x$ approaches zero from the positive side is essential for establishing limits like $\lim_{x \to 0^{+}} \sqrt{x} = 0$. The nature of $\sqrt{x}$ ensures that as $x$ gets extremely small, $\sqrt{x}$ also becomes extremely close to zero, satisfying the limit condition.

Limit Approaching Zero

The concept of a limit approaching zero is a specific case of the general limit concept, focusing on what happens to a function as the input moves very close to zero. It is especially significant in calculus as it helps understand the behavior of functions near points of interest.
In our exercise, we explore the limit of the square root function as its input, $x$, approaches zero from the positive side. This is noted as $\lim_{x \to 0^{+}} \sqrt{x} = 0$. What this means is:

As $x$ gets increasingly near to zero (but remains positive),
The value of $\sqrt{x}$ becomes smaller and tends towards zero.

This scenario illustrates the core idea of limits in calculus, where approaching does not mean reaching. The function approaches the limit value without necessarily attaining it. This principle is what helps in providing a meticulous understanding of function behavior and is foundational in solving more complex calculus problems.

Problem 26

Other exercises in this chapter

Problem 25

Find the limits. $\lim _{n \rightarrow \infty} \frac{n}{\sqrt{n^{2}+1}}$

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Problem 26

In Problems 19-28, use a calculator to find the indicated limit. Use a graphing calculator to plot the function near the limit point. $$ \lim _{t \rightarrow 0}

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Problem 26

In Problems 24-35, at what points, if any, are the functions discontinuous? $$ h(\theta)=|\sin \theta+\cos \theta| $$

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Problem 26

Find the limits if $\lim _{x \rightarrow a} f(x)=3$ and $\lim _{x \rightarrow a} g(x)=-1$ (see Example 4$)$. $$\lim _{x \rightarrow a} \frac{2 f(x)-3 g(x)

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