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Chapter 2: Problem 2

Use theorems on limits to find the limit, if it exists. $$ \lim _{x \rightarrow 15} \sqrt{2} $$

Short Answer

Expert verified
The limit is \( \sqrt{2} \).

Step by step solution

01

Identify the Expression

We need to find the limit of a constant function as it approaches a certain point. Here, the function is \( f(x) = \sqrt{2} \).
02

Recall Limit Property of Constant Functions

The limit of a constant function \( f(x) = c \) as \( x \) approaches any number \( a \) is the constant \( c \). This is because the value of the function does not change with \( x \).
03

Apply the Theorem

Since \( \sqrt{2} \) is a constant, the limit is simply the constant itself regardless of the value of \( x \) to which it is approaching. So, \( \lim_{x \to 15} \sqrt{2} = \sqrt{2} \).

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Key Concepts

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

Limit Laws
Understanding limit laws is crucial when studying calculus. Limit laws are a set of rules that help evaluate the limits of functions systematically and accurately. They form the backbone of solving limit-related problems in calculus, providing a framework that clarifies how limits behave in various situations.
These laws include the sum law, product law, quotient law, and power law, among others. They allow us to break down complicated expressions into simpler parts where limits can be calculated independently.
One important aspect of limit laws is their application to constant functions, where the behavior is often straightforward, as we'll see in the next sections.
Constant Function
A constant function is a type of function that assigns the same value to every input. In simpler terms, regardless of what value you plug into the function, it will always output the same number. For example, the function given in the problem, \( f(x) = \sqrt{2} \), always outputs \( \sqrt{2} \) no matter what \( x \) is.
Working with constant functions is typically straightforward because their value doesn't change. This constancy makes them easy to analyze, especially when exploring limits. With limits, no matter what \( x \) approaches, the function's output remains the same constant value. This property is very handy and simplifies calculations.
Calculus Limits
Limits are a fundamental concept in calculus used to describe the behavior of a function as its input approaches a particular point. They are essential in defining derivatives and integrals, which are the core building blocks of calculus.
When dealing with limits, one key property is understanding what happens to a function as the variable approaches a specific number. In the case of constant functions, this process is simplified, as the output does not change with different inputs.
Calculus limits allow us to tackle more complex changes in functions. However, for constant functions like \( f(x) = \sqrt{2} \), the limit is simply the value of the constant itself, showing the power of simplicity in limit laws.

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Most popular questions from this chapter

Give an example of a function \(f\) that is defined at \(a\) such that \(\lim _{x \rightarrow a} f(x)\) exists and \(\lim _{x \rightarrow a} f(x) \neq f(a)\).

Let []\(]\) denote the greatest integer function and \(n\) an arbitrary integer. Find (a) \(\lim _{x \rightarrow n^{-}} f(x)\) (b) \(\lim _{x \rightarrow n^{+}} f(x)\) $$ f(x)=-[-x] $$

Find each limit, if it exists: (a) \(\lim _{x \rightarrow a^{-}} f(x)\) (b) \(\lim _{x \rightarrow a^{+}} f(x)\) (c) \(\lim _{x \rightarrow a} f(x)\) $$ f(x)=x^{2 / 3} ; \quad a=-8 $$

Use the sandwich theorem to verify the limit. $$ \begin{aligned} &\lim _{x \rightarrow 0} x \sin (1 / x)=0\\\ &\text { (Hint: Use } f(x)=-|x| \text { and } g(x)=|x| \text { .) } \end{aligned} $$

Use theorems on limits to find the limit, if it exists. $$ \lim _{x \rightarrow 4} \sqrt[3]{x^{2}-5 x-4} $$
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