practice exercises\n17 - 36. limits at infinity determine the following limits.\n17. $lim_{\theta\rightarrowi…

Explanation:

Step1: Recall the range of cosine function

We know that $- 1\leqslant\cos\theta\leqslant1$ for all $\theta$. Then $\left|\frac{\cos\theta}{\theta^{2}}\right|\leqslant\frac{1}{\theta^{2}}$.

Step2: Apply the Squeeze - Theorem

We know that $\lim_{\theta\rightarrow\infty}\frac{-1}{\theta^{2}} = 0$ and $\lim_{\theta\rightarrow\infty}\frac{1}{\theta^{2}}=0$. By the Squeeze - Theorem, $\lim_{\theta\rightarrow\infty}\frac{\cos\theta}{\theta^{2}} = 0$.

Answer:

$0$

Explanation:

Step1: Split the fraction

$\lim_{t\rightarrow\infty}\frac{5t^{2}+t\sin t}{t^{2}}=\lim_{t\rightarrow\infty}\left(5 + \frac{\sin t}{t}\right)$.

Step2: Analyze the limit of each term

We know that $\lim_{t\rightarrow\infty}5 = 5$, and since $-1\leqslant\sin t\leqslant1$, then $\left|\frac{\sin t}{t}\right|\leqslant\frac{1}{t}$. Also, $\lim_{t\rightarrow\infty}\frac{1}{t}=0$, so $\lim_{t\rightarrow\infty}\frac{\sin t}{t}=0$ by the Squeeze - Theorem.

Step3: Combine the limits

$\lim_{t\rightarrow\infty}\left(5+\frac{\sin t}{t}\right)=5 + 0=5$.

Answer:

$5$

Explanation:

Step1: Recall the range of cosine function

We know that $-1\leqslant\cos(x^{5})\leqslant1$. Then $\left|\frac{\cos(x^{5})}{\sqrt{x}}\right|\leqslant\frac{1}{\sqrt{x}}$.

Step2: Apply the Squeeze - Theorem

Since $\lim_{x\rightarrow\infty}\frac{1}{\sqrt{x}} = 0$, by the Squeeze - Theorem, $\lim_{x\rightarrow\infty}\frac{\cos(x^{5})}{\sqrt{x}}=0$.

Answer:

$0$

Explanation:

Step1: Analyze the limit of each term

We know that $\lim_{x\rightarrow-\infty}5 = 5$, $\lim_{x\rightarrow-\infty}\frac{100}{x}=0$ (because as $x\rightarrow-\infty$, the denominator gets larger in magnitude and the fraction approaches 0), and since $- 1\leqslant\sin(x^{3})\leqslant1$, then $\left|\frac{\sin^{4}(x^{3})}{x^{2}}\right|\leqslant\frac{1}{x^{2}}$, and $\lim_{x\rightarrow-\infty}\frac{1}{x^{2}} = 0$.

Step2: Combine the limits

$\lim_{x\rightarrow-\infty}\left(5+\frac{100}{x}+\frac{\sin^{4}(x^{3})}{x^{2}}\right)=5 + 0+0 = 5$.

Answer:

$5$

Explanation:

Step1: Consider the leading - term

For the polynomial $y = 3x^{12}-9x^{7}$, as $x\rightarrow\infty$, the leading - term is $3x^{12}$ (because the degree of $x^{12}$ is higher than the degree of $x^{7}$).

Step2: Determine the limit

Since the coefficient of the leading - term $3>0$ and the degree $n = 12$ (even), $\lim_{x\rightarrow\infty}(3x^{12}-9x^{7})=\infty$.

Answer:

$\infty$

Explanation:

Step1: Consider the leading - term

For the polynomial $y = 3x^{7}+x^{2}$, as $x\rightarrow-\infty$, the leading - term is $3x^{7}$ (because the degree of $x^{7}$ is higher than the degree of $x^{2}$).

Step2: Determine the limit

Since the coefficient of the leading - term $3>0$ and the degree $n = 7$ (odd), $\lim_{x\rightarrow-\infty}(3x^{7}+x^{2})=-\infty$.

Answer:

$-\infty$

Explanation:

Step1: Consider the leading - term

For the polynomial $y=-3x^{16}+2$, as $x\rightarrow-\infty$, the leading - term is $-3x^{16}$ (because the degree of $x^{16}$ is higher than the constant term).

Step2: Determine the limit

Since the coefficient of the leading - term $-3<0$ and the degree $n = 16$ (even), $\lim_{x\rightarrow-\infty}(-3x^{16}+2)=-\infty$.

Answer:

$-\infty$

Explanation:

Step1: Analyze the limit of each term

We know that $\lim_{x\rightarrow-\infty}2x^{-8}=\lim_{x\rightarrow-\infty}\frac{2}{x^{8}} = 0$ (because as $x\rightarrow-\infty$, $x^{8}\rightarrow\infty$) and $\lim_{x\rightarrow-\infty}4x^{3}=-\infty$ (since the coefficient $4>0$ and the degree $n = 3$ (odd)).

Step2: Combine the limits

$\lim_{x\rightarrow-\infty}(2x^{-8}+4x^{3})=-\infty$.

Answer:

$-\infty$

Explanation:

Step1: Divide both numerator and denominator by the highest - power of $x$ in the denominator

Divide $\frac{14x^{3}+3x^{2}-2x}{21x^{3}+x^{2}+2x + 1}$ by $x^{3}$: $\lim_{x\rightarrow\infty}\frac{14x^{3}+3x^{2}-2x}{21x^{3}+x^{2}+2x + 1}=\lim_{x\rightarrow\infty}\frac{14+\frac{3}{x}-\frac{2}{x^{2}}}{21+\frac{1}{x}+\frac{2}{x^{2}}+\frac{1}{x^{3}}}$.

Step2: Determine the limit of each term

As $x\rightarrow\infty$, $\lim_{x\rightarrow\infty}\frac{3}{x}=0$, $\lim_{x\rightarrow\infty}\frac{2}{x^{2}} = 0$, $\lim_{x\rightarrow\infty}\frac{1}{x}=0$, $\lim_{x\rightarrow\infty}\frac{2}{x^{2}} = 0$, $\lim_{x\rightarrow\infty}\frac{1}{x^{3}} = 0$.

Step3: Combine the limits

$\lim_{x\rightarrow\infty}\frac{14+\frac{3}{x}-\frac{2}{x^{2}}}{21+\frac{1}{x}+\frac{2}{x^{2}}+\frac{1}{x^{3}}}=\frac{14 + 0-0}{21+0 + 0+0}=\frac{2}{3}$.

Answer:

$\frac{2}{3}$

Explanation:

Step1: Divide both numerator and denominator by the highest - power of $x$ in the denominator

Divide $\frac{9x^{3}+x^{2}-5}{3x^{4}+4x^{2}}$ by $x^{4}$: $\lim_{x\rightarrow\infty}\frac{9x^{3}+x^{2}-5}{3x^{4}+4x^{2}}=\lim_{x\rightarrow\infty}\frac{\frac{9}{x}+\frac{1}{x^{2}}-\frac{5}{x^{4}}}{3+\frac{4}{x^{2}}}$.

Step2: Determine the limit of each term

As $x\rightarrow\infty$, $\lim_{x\rightarrow\infty}\frac{9}{x}=0$, $\lim_{x\rightarrow\infty}\frac{1}{x^{2}} = 0$, $\lim_{x\rightarrow\infty}\frac{5}{x^{4}} = 0$, $\lim_{x\rightarrow\infty}\frac{4}{x^{2}} = 0$.

Step3: Combine the limits

$\lim_{x\rightarrow\infty}\frac{\frac{9}{x}+\frac{1}{x^{2}}-\frac{5}{x^{4}}}{3+\frac{4}{x^{2}}}=\frac{0 + 0-0}{3+0}=0$.

Answer:

$0$

Explanation:

Step1: Divide both numerator and denominator by the highest - power of $x$ in the denominator

Divide $\frac{3x^{2}+3x}{x + 1}$ by $x$: $\lim_{x\rightarrow-\infty}\frac{3x^{2}+3x}{x + 1}=\lim_{x\rightarrow-\infty}\frac{3x + 3}{1+\frac{1}{x}}$.

Step2: Determine the limit of each term

As $x\rightarrow-\infty$, $\lim_{x\rightarrow-\infty}\frac{1}{x}=0$, and $\lim_{x\rightarrow-\infty}(3x + 3)=-\infty$ (since the coefficient of $x$ is $3>0$ and $x\rightarrow-\infty$).

Step3: Combine the limits

$\lim_{x\rightarrow-\infty}\frac{3x + 3}{1+\frac{1}{x}}=-\infty$.

Answer:

$-\infty$

Explanation:

Step1: Divide both numerator and denominator by the highest - power of $x$ in the denominator

Divide $\frac{x^{4}+7}{x^{5}+x^{2}-x}$ by $x^{5}$: $\lim_{x\rightarrow\infty}\frac{x^{4}+7}{x^{5}+x^{2}-x}=\lim_{x\rightarrow\infty}\frac{\frac{1}{x}+\frac{7}{x^{5}}}{1+\frac{1}{x^{3}}-\frac{1}{x^{4}}}$.

Step2: Determine the limit of each term

As $x\rightarrow\infty$, $\lim_{x\rightarrow\infty}\frac{1}{x}=0$, $\lim_{x\rightarrow\infty}\frac{7}{x^{5}} = 0$, $\lim_{x\rightarrow\infty}\frac{1}{x^{3}} = 0$, $\lim_{x\rightarrow\infty}\frac{1}{x^{4}} = 0$.

Step3: Combine the limits

$\lim_{x\rightarrow\infty}\frac{\frac{1}{x}+\frac{7}{x^{5}}}{1+\frac{1}{x^{3}}-\frac{1}{x^{4}}}=\frac{0 + 0}{1+0 - 0}=0$.

Answer:

$0$