Example: Using the Squeeze Theorem
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Consider the function \(f(x) = x^2\sin\frac{1}{x}\). We want to find the limit as \(x\) approaches 0.
- The sine function \(\sin \theta\) oscillates between -1 and 1 for all values of \(\theta\). As \(x\) approaches 0, \(\frac{1}{x}\) becomes very large in magnitude, causing \(\sin \frac{1}{x}\) to oscillate rapidly between -1 and 1. Thus, we have: \( -1 \leq \sin\frac{1}{x} \leq 1 \)
- Next, we multiply the above inequality by \(x^2\) (which is always non-negative): \( -x^2 \leq x^2\sin\frac{1}{x} \leq x^2 \)
- Since \(x^2\) approaches 0 as \(x\) approaches 0, we can use the Squeeze Theorem to find the limit: \[ \lim_{{x\to 0}} (-x^2) \leq \lim_{{x\to 0}} (x^2\sin\frac{1}{x}) \leq \lim_{{x\to 0}} (x^2) \]
- As \(x\) approaches 0, both \((-x^2)\) and \(x^2\) tend to 0, so we have: \( 0 \leq \lim_{{x\to 0}} (x^2\sin\frac{1}{x}) \leq 0 \)
- By the Squeeze Theorem, since the limit of \((-x^2)\) and \(x^2\) are both 0, the limit of \(x^2\sin\frac{1}{x}\) as \(x\) approaches 0 is also 0: \( \lim_{{x\to 0}} (x^2\sin\frac{1}{x}) = 0 \)
Example: Using the Squeeze Theorem
Consider the function \(g(x) = x^3\cos\frac{1}{x}\). We want to find the limit as \(x\) approaches 0.
- The cosine function \(\cos \theta\) oscillates between -1 and 1 for all values of \(\theta\).
As \(x\) approaches 0, \(\frac{1}{x}\) becomes very large in magnitude, causing \(\cos \frac{1}{x}\) to oscillate rapidly between -1 and 1.
Thus, we have: \( -1 \leq \cos\frac{1}{x} \leq 1 \) - Next, we multiply the above inequality by \(|x^3|\) (the absolute value of \(x^3\)):
\( -|x^3| \leq x^3\cos\frac{1}{x} \leq |x^3| \) - Since \(|x^3|\) is non-negative and approaches 0 as \(x\) approaches 0, we can use the Squeeze Theorem to find the limit:
\[ \lim_{{x\to 0}} (-|x^3|) \leq \lim_{{x\to 0}} (x^3\cos\frac{1}{x}) \leq \lim_{{x\to 0}} (|x^3|) \] - As \(x\) approaches 0, both \((-|x^3|)\) and \(|x^3|\) tend to 0, so we have:
\( 0 \leq \lim_{{x\to 0}} (x^3\cos\frac{1}{x}) \leq 0 \) - By the Squeeze Theorem, since the limit of \((-|x^3|)\) and \(|x^3|\) are both 0, the limit of \(x^3\cos\frac{1}{x}\) as \(x\) approaches 0 is also 0:
\( \lim_{{x\to 0}} (x^3\cos\frac{1}{x}) = 0 \)
Example: Using the Squeeze Theorem
Consider the function \(h(x) = \sqrt{x}\sin\frac{1}{x}\). We want to find the limit as \(x\) approaches 0.
- The sine function \(\sin \theta\) oscillates between -1 and 1 for all values of \(\theta\).
As \(x\) approaches 0, \(\frac{1}{x}\) becomes very large in magnitude, causing \(\sin \frac{1}{x}\) to oscillate rapidly between -1 and 1.
Thus, we have: \( -1 \leq \sin\frac{1}{x} \leq 1 \) - Next, we multiply the above inequality by \(\sqrt{x}\) (which is always non-negative):
\( -\sqrt{x} \leq \sqrt{x}\sin\frac{1}{x} \leq \sqrt{x} \) - Since \(\sqrt{x}\) approaches 0 as \(x\) approaches 0, we can use the Squeeze Theorem to find the limit:
\[ \lim_{{x\to 0}} (-\sqrt{x}) \leq \lim_{{x\to 0}} (\sqrt{x}\sin\frac{1}{x}) \leq \lim_{{x\to 0}} (\sqrt{x}) \] - As \(x\) approaches 0, both \((-\sqrt{x})\) and \(\sqrt{x}\) tend to 0, so we have:
\( 0 \leq \lim_{{x\to 0}} (\sqrt{x}\sin\frac{1}{x}) \leq 0 \) - By the Squeeze Theorem, since the limit of \((-\sqrt{x})\) and \(\sqrt{x}\) are both 0, the limit of \(\sqrt{x}\sin\frac{1}{x}\) as \(x\) approaches 0 is also 0:
\( \lim_{{x\to 0}} (\sqrt{x}\sin\frac{1}{x}) = 0 \)
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Example: Using the Squeeze Theorem with Piecewise Function
Consider the function \(h(x) = \begin{cases} x^2\sin\frac{1}{x} & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases}\). We want to find the limit as \(x\) approaches 0.
- For \(x \neq 0\), the sine function \(\sin \theta\) oscillates between -1 and 1 for all values of \(\theta\).
As \(x\) approaches 0, \(\frac{1}{x}\) becomes very large in magnitude, causing \(\sin \frac{1}{x}\) to oscillate rapidly between -1 and 1.
Thus, we have: \( -x^2 \leq x^2\sin\frac{1}{x} \leq x^2 \) - Since \(x^2\) approaches 0 as \(x\) approaches 0, we can use the Squeeze Theorem to find the limit:
\[ \lim_{{x\to 0}} (-x^2) \leq \lim_{{x\to 0}} (x^2\sin\frac{1}{x}) \leq \lim_{{x\to 0}} (x^2) \] - As \(x\) approaches 0, both \((-x^2)\) and \(x^2\) tend to 0, so we have:
\( 0 \leq \lim_{{x\to 0}} (x^2\sin\frac{1}{x}) \leq 0 \) - By the Squeeze Theorem, since the limit of \((-x^2)\) and \(x^2\) are both 0, the limit of \(x^2\sin\frac{1}{x}\) as \(x\) approaches 0 is also 0:
\( \lim_{{x\to 0}} (x^2\sin\frac{1}{x}) = 0 \) - Additionally, for \(x = 0\), the function \(h(x)\) is defined as 0. Therefore, \( \lim_{{x\to 0}} h(x) = 0 \) (by direct substitution).
- Since the limits from both approaches are equal, we conclude that the overall limit of \(h(x)\) as \(x\) approaches 0 is 0:
\( \lim_{{x\to 0}} h(x) = 0 \)
Example: Using the Squeeze Theorem with Trigonometric Function
Consider the function \(g(x) = \frac{\sin x}{x}\). We want to find the limit as \(x\) approaches 0.
- As \(x\) approaches 0, \(\sin x\) approaches 0, and \(x\) also approaches 0.
- We can use the Squeeze Theorem to bound \(g(x)\) using the inequality \( -1 \leq \frac{\sin x}{x} \leq 1 \) since \(-1 \leq \sin x \leq 1\).
- Since the limit of both \(-1\) and \(1\) as \(x\) approaches 0 is 0, we can conclude that the limit of \(g(x)\) as \(x\) approaches 0 is also 0: \( \lim_{{x\to 0}} \frac{\sin x}{x} = 0 \)
Example: Find \( \lim_{{x\to 0}} x^2 \cos^2 \frac{1}{x} \) using the Squeeze Theorem.
As \(x\) approaches 0, both \(x^2\) and \(\cos^2 \frac{1}{x}\) vary. However, we know that \(\cos^2 \frac{1}{x}\) is always between 0 and 1 for all \(x\) due to the squaring operation on the range of \(\cos\) function.
We can establish a bound for \(x^2 \cos^2 \frac{1}{x}\) by considering the inequality \(0 \leq \cos^2 \frac{1}{x} \leq 1\) for all \(x\).
Since \(x^2\) is always non-negative (positive or zero) for all \(x\), multiplying the above inequality by \(x^2\) does not change the inequality: \(0 \leq x^2 \cos^2 \frac{1}{x} \leq x^2\) for all \(x\).
Now, as \(x\) approaches 0, both \(0\) and \(x^2\) tend to 0.
Taking the limits on both the LHS and the RHS of the inequality, we have:
\[ \lim_{{x\to 0}} 0 \leq \lim_{{x\to 0}} x^2 \cos^2 \frac{1}{x} \leq \lim_{{x\to 0}} x^2 \]Since the limits of both \(0\) and \(x^2\) as \(x\) approaches 0 are 0, the limit of \(x^2 \cos^2 \frac{1}{x}\) as \(x\) approaches 0 is also 0:
\[ \lim_{{x\to 0}} x^2 \cos^2 \frac{1}{x} = 0 \]