Math / Calculus

Question4. It is possible to use the Squeeze Theorem to prove that $\lim _{x \rightarrow 0}\left(x \cdot \sin \left(\frac{1}{x}\right)\right)=0$ . Why couldn't we obtain the same result by writing $\begin{aligned} \lim _{x \rightarrow 0}\left(x \cdot \sin \left(\frac{1}{x}\right)\right) & =\lim _{x \rightarrow 0} x \cdot \lim _{x \rightarrow 0}\left(\sin \left(\frac{1}{x}\right)\right) \\ & =0 \cdot \lim _{x \rightarrow 0}\left(\sin \left(\frac{1}{x}\right)\right)=0 ? \end{aligned}$

Studdy Solution

STEP 1

1. The limit in question is $\lim_{x \to 0} \left(x \cdot \sin \left(\frac{1}{x}\right)\right)$ .
2. The Squeeze Theorem is applicable when we can find two functions that "squeeze" the given function and have the same limit at a particular point.
3. The limit $\lim_{x \to 0} x$ is straightforward and equals $0$ .
4. The function $\sin\left(\frac{1}{x}\right)$ oscillates between $-1$ and $1$ as $x \to 0$ .
5. The product of two limits $\lim_{x \to 0} (f(x) \cdot g(x))$ is generally $\lim_{x \to 0} f(x) \cdot \lim_{x \to 0} g(x)$ provided both limits exist and are finite.

STEP 2

1. Analyze why the direct limit approach fails.
2. Use the Squeeze Theorem to find the limit.

STEP 3

Consider the expression $\lim_{x \to 0} x \cdot \sin \left(\frac{1}{x}\right)$ .
To directly split the limit into two parts: $\lim_{x \to 0} \left(x \cdot \sin \left(\frac{1}{x}\right)\right) = \lim_{x \to 0} x \cdot \lim_{x \to 0} \sin \left(\frac{1}{x}\right)$

STEP 4

Evaluate $\lim_{x \to 0} \sin \left(\frac{1}{x}\right)$ .
Since $\sin \left(\frac{1}{x}\right)$ oscillates between $-1$ and $1$ without settling to a particular value as $x \to 0$ , the limit does not exist.

STEP 5

Since $\lim_{x \to 0} \sin \left(\frac{1}{x}\right)$ does not exist, we cannot use the property $\lim_{x \to 0} (f(x) \cdot g(x)) = \lim_{x \to 0} f(x) \cdot \lim_{x \to 0} g(x)$ .
Thus, the direct approach fails.

STEP 6

To use the Squeeze Theorem, identify functions that bound $x \cdot \sin \left(\frac{1}{x}\right)$ .
Note that $-1 \leq \sin \left(\frac{1}{x}\right) \leq 1$ for all $x \neq 0$ .

STEP 7

Multiply this inequality by $x$ , giving:
$-x \leq x \cdot \sin \left(\frac{1}{x}\right) \leq x$

STEP 8

Consider the limits of the bounding functions as $x \to 0$ :
$\lim_{x \to 0} (-x) = 0 \quad \text{and} \quad \lim_{x \to 0} x = 0$

STEP 9

Apply the Squeeze Theorem:
Since $-x \leq x \cdot \sin \left(\frac{1}{x}\right) \leq x$ and both $-x$ and $x$ tend to $0$ as $x \to 0$ , it follows that:
$\lim_{x \to 0} x \cdot \sin \left(\frac{1}{x}\right) = 0$

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Studdy Solution

STEP 1

STEP 2

1. Analyze why the direct limit approach fails.
2. Use the Squeeze Theorem to find the limit.

STEP 3

Consider the expression $\lim_{x \to 0} x \cdot \sin \left(\frac{1}{x}\right)$ .
To directly split the limit into two parts: $\lim_{x \to 0} \left(x \cdot \sin \left(\frac{1}{x}\right)\right) = \lim_{x \to 0} x \cdot \lim_{x \to 0} \sin \left(\frac{1}{x}\right)$

STEP 4

Evaluate $\lim_{x \to 0} \sin \left(\frac{1}{x}\right)$ .
Since $\sin \left(\frac{1}{x}\right)$ oscillates between $-1$ and $1$ without settling to a particular value as $x \to 0$ , the limit does not exist.

STEP 5

Since $\lim_{x \to 0} \sin \left(\frac{1}{x}\right)$ does not exist, we cannot use the property $\lim_{x \to 0} (f(x) \cdot g(x)) = \lim_{x \to 0} f(x) \cdot \lim_{x \to 0} g(x)$ .
Thus, the direct approach fails.

STEP 6

To use the Squeeze Theorem, identify functions that bound $x \cdot \sin \left(\frac{1}{x}\right)$ .
Note that $-1 \leq \sin \left(\frac{1}{x}\right) \leq 1$ for all $x \neq 0$ .

STEP 7

Multiply this inequality by $x$ , giving:
$-x \leq x \cdot \sin \left(\frac{1}{x}\right) \leq x$

STEP 8

Consider the limits of the bounding functions as $x \to 0$ :
$\lim_{x \to 0} (-x) = 0 \quad \text{and} \quad \lim_{x \to 0} x = 0$

STEP 9

Apply the Squeeze Theorem:
Since $-x \leq x \cdot \sin \left(\frac{1}{x}\right) \leq x$ and both $-x$ and $x$ tend to $0$ as $x \to 0$ , it follows that:
$\lim_{x \to 0} x \cdot \sin \left(\frac{1}{x}\right) = 0$

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Studdy Solution

STEP 1

STEP 2

1. Analyze why the direct limit approach fails.2. Use the Squeeze Theorem to find the limit.

STEP 3

STEP 4

Evaluate lim⁡x→0sin⁡(1x)\lim_{x \to 0} \sin \left(\frac{1}{x}\right)limx→0​sin(x1​).Since sin⁡(1x)\sin \left(\frac{1}{x}\right)sin(x1​) oscillates between −1-1−1 and 111 without settling to a particular value as x→0x \to 0x→0, the limit does not exist.

STEP 5

STEP 6

To use the Squeeze Theorem, identify functions that bound x⋅sin⁡(1x)x \cdot \sin \left(\frac{1}{x}\right)x⋅sin(x1​).Note that −1≤sin⁡(1x)≤1-1 \leq \sin \left(\frac{1}{x}\right) \leq 1−1≤sin(x1​)≤1 for all x≠0x \neq 0x=0.

STEP 7

Multiply this inequality by xxx, giving:−x≤x⋅sin⁡(1x)≤x-x \leq x \cdot \sin \left(\frac{1}{x}\right) \leq x−x≤x⋅sin(x1​)≤x

STEP 8

Consider the limits of the bounding functions as x→0x \to 0x→0:lim⁡x→0(−x)=0andlim⁡x→0x=0\lim_{x \to 0} (-x) = 0 \quad \text{and} \quad \lim_{x \to 0} x = 0x→0lim​(−x)=0andx→0lim​x=0

STEP 9

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1. Analyze why the direct limit approach fails.
2. Use the Squeeze Theorem to find the limit.

Evaluate $\lim_{x \to 0} \sin \left(\frac{1}{x}\right)$ .
Since $\sin \left(\frac{1}{x}\right)$ oscillates between $-1$ and $1$ without settling to a particular value as $x \to 0$ , the limit does not exist.

To use the Squeeze Theorem, identify functions that bound $x \cdot \sin \left(\frac{1}{x}\right)$ .
Note that $-1 \leq \sin \left(\frac{1}{x}\right) \leq 1$ for all $x \neq 0$ .

Multiply this inequality by $x$ , giving:
$-x \leq x \cdot \sin \left(\frac{1}{x}\right) \leq x$

Consider the limits of the bounding functions as $x \to 0$ :
$\lim_{x \to 0} (-x) = 0 \quad \text{and} \quad \lim_{x \to 0} x = 0$