Free Math Worksheets

- Exclude inputs that make the denominator zero. - Use the zeros of sine. - Bound the absolute value using \(|\sin(x)|\le1\). - Solve the simpler inequality involving \(\frac{5}{x}\).

1. The denominator cannot be zero, so the domain is \(\mathbb{R}\setminus\{0\}\). 2. The factor \(\frac{5}{x}\) is never zero. Therefore, zeros occur where \(\sin(x)=0\), at \(x=k\pi\) for \(k\in\mathbb{Z}\setminus\{0\}\). 3. Since \(|\sin(x)|\le1\), \(\left|\frac{5\sin(x)}{x}\right|\le\frac{5}{x}\) for \(x>0\). Because \(\frac{5}{x}\to0\), the squeeze theorem gives \(f(x)\to0\). 4. To guarantee \(|f(x)|<0.01\), it is enough to require \(\frac{5}{x}<0.01\). This gives \(x>500\), so \(x_0=500\) works.

a) \(\mathbb{R}\setminus\{0\}\) b) \(x=k\pi\), where \(k\in\mathbb{Z}\setminus\{0\}\) c) \(0\) d) \(x_0=500\)

- Use \(|\cos(x)|\le1\). - Bound the variable fraction above and below. - Replace the oscillating numerator with its maximum possible absolute value to find a sufficient threshold.

1. The denominator is zero at \(x=0\), so the domain is \(\mathbb{R}\setminus\{0\}\). 2. For \(x\ge1\), \(\left|\frac{\cos(x)}{x^2}\right|\le\frac{1}{x^2}\le1\). Therefore, \(g(x)\ge2-1=1>0\), so there are no zeros on \([1, \infty)\). 3. Since \(|\cos(x)|\le1\) and \(x^2\to\infty\), the squeeze theorem gives \(\frac{\cos(x)}{x^2}\to0\) as \(x\to-\infty\). Thus, \(L=2\). 4. To guarantee \(|g(x)-2|<0.0025\), it is enough to require \(\frac{1}{x^2}<0.0025\). For positive \(x\), this gives \(x>20\), so \(x_0=20\) works.

a) \(\mathbb{R}\setminus\{0\}\) b) \(g(x)\ge1>0\) for \(x\ge1\). c) \(L=2\) d) \(x_0=20\)

- Bound the sine factor by its maximum absolute value. - Multiply the bound by \(|x-2|\). - Apply the squeeze theorem.

1. Since \(\left|\sin\left(\frac{1}{x-2}\right)\right|\le1\), \(|f(x)|\le|x-2|\). 2. Equivalently, \(-|x-2|\le f(x)\le|x-2|\). 3. Both bounding functions approach \(0\) as \(x\to2\). 4. By the squeeze theorem, \(\lim_{x\to2}f(x)=0\).

The limit exists and equals \(0\).

- Split the fraction into two terms. - Use the bounds on cosine. - Apply the squeeze theorem to the oscillating term.

1. Split the fraction: \(g(x)=3+\frac{\cos(x^2)}{x}\). 2. Since \(-1\le\cos(x^2)\le1\), \(-\frac{1}{x}\le\frac{\cos(x^2)}{x}\le\frac{1}{x}\) for \(x>0\). 3. Both bounds approach \(0\) as \(x\to\infty\), so the squeeze theorem gives \(\frac{\cos(x^2)}{x}\to0\). 4. Therefore, \(g(x)\to3\).

\(\lim_{x\to\infty}g(x)=3\)

- Use the fact that \(-1\le\sin(x)\le1\). - Create upper and lower bounds for \(\frac{8\sin(x)}{x}\). - Apply the Squeeze Theorem, then add the constant \(2\).

1. The graph oscillates around \(y=2\), and the size of the oscillations decreases as \(x\) increases. 2. Since \(-1\le\sin(x)\le1\), for \(x>0\) we have \(-\frac{8}{x}\le\frac{8\sin(x)}{x}\le\frac{8}{x}\). 3. Both bounding functions approach \(0\) as \(x\to\infty\). By the Squeeze Theorem, \(\lim_{x\to\infty}\frac{8\sin(x)}{x}=0\). 4. Therefore, \(\lim_{x\to\infty}f(x)=2\), and the horizontal asymptote is \(y=2\).

\(\lim_{x\to\infty}f(x)=2\), and the horizontal asymptote is \(y=2\).