Problem 453

Question

\begin{aligned} &\text { Sketch the graph of the function }\\\ &f(x)=\lim _{n \rightarrow \infty} \sqrt[n]{1+x^{n}}, \quad x \geq 0 \end{aligned}

Step-by-Step Solution

Verified

Answer

The function $ f(x) = \lim_{n \rightarrow \infty} \sqrt[n]{1+x^{n}} $, with $ x \geq 0 $, is equal to 1. Therefore, the graph of the function is a horizontal line at $ y = 1 $.

1Step 1: Consider the limit

We know that for $ x \geq 0 $, $ x^n $ will always be greater than or equal to 1. This makes the expression inside the nth root always greater than or equal to 1, and the nth root of a number greater than or equal to 1 is always less than or equal to the number itself. Therefore, we can say that $ 1 \leq f(x) \leq 1+x^n $ for any $ x \geq 0 $.

2Step 2: Apply the Squeeze theorem

The Squeeze (or Sandwich) theorem says that if we have three functions, $ f(x) $, $ g(x) $, and $ h(x) $, and $ f(x) \leq g(x) \leq h(x) $, then if the limit as $ x \rightarrow a $ of $ f(x) $ and $ h(x) $ are both $ L $, the limit of $ g(x) $ as $ x \rightarrow a $ is also $ L $. Here, we're letting $ f(x) = 1 $, $ g(x) = \sqrt[n]{1+x^{n}} $, and $ h(x) = 1+x^n $. Taking the limit as $ n \rightarrow \infty $ leaves us with $ L = 1 $. Therefore, by the Squeeze theorem, $ \lim_{n \rightarrow \infty} \sqrt[n]{1+x^{n}} = 1 $.

3Step 3: Sketch the Graph

Since the limit as $ n \rightarrow \infty $ of $ f(x) $, given $ x \geq 0 $, is 1, this means the function $ f(x) $ is equal to 1 for all $ x \geq 0 $. The graph of $ f(x) = 1 $ is simply a horizontal line at $ y=1 $.

Key Concepts

Squeeze Theoremnth Root PropertiesGraph Sketching

Squeeze Theorem

The Squeeze Theorem, also known as the Sandwich Theorem, is a crucial concept in calculus, particularly when dealing with limits that are difficult to evaluate using conventional methods.

Imagine you have a function, let's call it $g(x)$, which is 'squeezed' between two other functions $f(x)$ and $h(x)$. Now, if both $f(x)$ and $h(x)$ approach the same limit as $x$ approaches a certain value, then $g(x)$ is forced to approach that same limit as well. In mathematical terms, if $\lim_{x \rightarrow a} f(x) = L$ and $\lim_{x \rightarrow a} h(x) = L$, and $f(x) \leq g(x) \leq h(x)$ for all $x$ in an interval containing $a$, then $\lim_{x \rightarrow a} g(x) = L$.

In our exercise, $f(x)$ is the constant function $1$, and $h(x)$ is the function $1+x^n$, which also approaches $1$ as $n$ approaches infinity. Because $g(x)$, the nth root function, lies between these two, its limit must be $1$ as well. This application is a prime example of how the Squeeze Theorem helps us find limits that are not immediately obvious.

nth Root Properties

Understanding nth root properties is vital when working with complex limits, as in our exercise scenario. The nth root of a number $a$, written as $\sqrt[n]{a}$, is a value that, when raised to the power $n$, gives back the number $a$.

Here are some essential properties of nth roots:

For any positive number $a$, $\sqrt[n]{a} \geq 0$.
If $a = b^n$, then $\sqrt[n]{a} = b$ and $\sqrt[n]{b^n} = b$ provided that $b \geq 0$.
If $a\geq b\geq 0$, then $\sqrt[n]{a} \geq \sqrt[n]{b}$.
As $n$ becomes very large, $\sqrt[n]{a}$ approaches $1$ if $a > 1$, and stays at $0$ if $a = 0$.

In the context of the given exercise, we use the property that the nth root of $1+x^n$ will always be less than or equal to $1+x^n$ itself, which helps us establish the upper bound for our Squeeze Theorem application.

Graph Sketching

Graph sketching is a valuable skill in calculus that enables students to visualize the behavior of functions. A well-drawn graph can reveal a function's continuity, limits, and points of interest like intercepts and asymptotes.

When sketching the graph of a function, one must consider several aspects:

Intercepts:
Where does the function cross the axes?
Asymptotes:
Are there any values the function approaches but never reaches?
Intervals of increase or decrease:
In which sections is the function going upwards or downwards?
Curvature:
Is the graph concave up or down?
Continuity and differentiability:
Are there any points where the function is not continuous or differentiable?

In our particular function $f(x) = \lim_{n \rightarrow \infty} \sqrt[n]{1+x^{n}}$, the limit as $n$ goes to infinity for all $x \geq 0$ simplifies to $1$. This simplification indicates that for all non-negative values of $x$, the function takes on the value $1$. Therefore, the graph is a flat, horizontal line at $y=1$ for $x \geq 0$. This horizontal line represents the function's consistent outcome, regardless of the changes in $x$, as long as $x$ remains non-negative.

Problem 452

Problem 454