Squeeze Theorem to Prove Limit of x/sin(x) as x approaches 0

Math Problem Statement

We want to find

[\displaystyle\lim_{x\to 0}\dfrac{x}{\text{sin}(x)}]. Direct substitution and other algebraic methods don't seem to work. Looking at the graph of

[\blueD{f(x)=\dfrac{x}{\text{sin}(x)}}], we can estimate that the limit is equal to [1]. Function f is graphed. The x-axis goes from negative 3 to 3. The graph consists of a U-shaped curve. The curve starts in quadrant 2, moves downward through about (negative 1.5, 1.5) to an open circle at (0, 1), moves upward through about (1.5, 1.5) and ends in quadrant 1. [\small{1}] [\small{2}] [\small{\llap{-}2}] [\small{1}] [\small{2}] [\small{3}] [\small{4}] [y] [x] [\small -1] [\blueD f] To prove that

[\displaystyle\lim_{x\to 0}\dfrac{x}{\text{sin}(x)}=1], we can use the squeeze theorem. Delu suggested that we use the functions [\goldD{g(x)=1}] and [\maroonD{h(x)=3-2\text{cos}(x)}] in order to apply the squeeze theorem. Functions f, g, and h are graphed. The x-axis goes from negative 3 to 3. The graph of function f consists of a U-shaped curve. The curve starts in quadrant 2, moves downward through about (negative 1.5, 1.5) to an open circle at (0, 1), moves upward through about (1.5, 1.5), and ends in quadrant 1. The graph of function h consists of a curve. The curve starts about (negative 3, 5), moves downward gradually, then more steeply, then gradually to (0, 1), then moves upward gradually, then more steeply, then gradually, and ends at about (3, 5). The graph of function g is a horizontal line, moving through (0, 1). [\small{1}] [\small{2}] [\small{\llap{-}2}] [\small{1}] [\small{2}] [\small{3}] [\small{4}] [y] [x] [\small -1] [\blueD f] [\goldD g] [\maroonD h] Does Delu's suggestion seem to be correct? Choose 1 answer: Choose 1 answer: (Choice A) Yes, Delu's suggestion seems to be correct. A Yes, Delu's suggestion seems to be correct. (Choice B) No, Delu's suggestion is incorrect because it's not true that the output of one function is always below [f(x)] and the output of the other function is always above it for [x]-values near [0]. B No, Delu's suggestion is incorrect because it's not true that the output of one function is always below [f(x)] and the output of the other function is always above it for [x]-values near [0]. (Choice C) No, Delu's suggestion is incorrect because it's not true that the limits of [\goldD{g}] and [\maroonD{h}] are both equal to [1]. C No, Delu's suggestion is incorrect because it's not true that the limits of [\goldD{g}] and [\maroonD{h}] are both equal to [1].

Solution

The goal is to prove that:

$\lim_{x \to 0} \frac{x}{\sin(x)} = 1$

Using the Squeeze Theorem, we need to find two functions, $g(x)$ and $h(x)$ , such that:

$g(x) \leq f(x) \leq h(x)$ for values of $x$ near 0.
The limits of $g(x)$ and $h(x)$ as $x \to 0$ are both equal to 1.

From the graph, the function $f(x) = \frac{x}{\sin(x)}$ is a U-shaped curve, with the open circle at $(0, 1)$ , indicating that it approaches 1 as $x \to 0$ .

Now, analyzing Delu's suggestion:

$g(x) = 1$ , which is a horizontal line at $y = 1$ .
$h(x) = 3 - 2\cos(x)$ , which is a curve that approaches 1 as $x \to 0$ because $\cos(0) = 1$ , and thus $h(0) = 3 - 2(1) = 1$ .

For values near $x = 0$ , the function $g(x)$ is indeed always below $f(x)$ , and $h(x)$ is always above $f(x)$ , and both limits of $g(x)$ and $h(x)$ as $x \to 0$ approach 1. Therefore, Delu's suggestion to use $g(x) = 1$ and $h(x) = 3 - 2\cos(x)$ is correct, and by the Squeeze Theorem, we can conclude that:

$\lim_{x \to 0} \frac{x}{\sin(x)} = 1$

Thus, the correct choice is:

(Choice A) Yes, Delu's suggestion seems to be correct.

Would you like further clarification on the Squeeze Theorem or how it applies here?

Here are some related questions you might find interesting:

How does the Squeeze Theorem work in general for proving limits?
What happens if we try direct substitution in this limit without the Squeeze Theorem?
Can the Squeeze Theorem be used for limits involving other trigonometric functions?
What are some other common limits involving trigonometric functions that use the Squeeze Theorem?
How do we derive the limit $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$ using similar reasoning?

Tip: When applying the Squeeze Theorem, always check that the lower and upper functions bound the target function from below and above in the vicinity of the point you're evaluating.

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Math Problem Analysis

Mathematical Concepts

Limits
Trigonometry
Squeeze Theorem

Formulas

\lim_{x \to 0} \dfrac{x}{\sin(x)} = 1
g(x) = 1
h(x) = 3 - 2\cos(x)

Theorems

Squeeze Theorem

Suitable Grade Level

Grades 11-12

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