Question

1. a. (y = e^{x - 3}); over (3,7) b. (y=sqrt{6 - 3x}); over (0,2) 2. find the limit. a. (lim_{x

ightarrow - 1}(2x^{2}-3x)) b. (lim_{x
ightarrow3}\frac{x^{2}-2x - 3}{x^{2}-5x + 6}) c. (lim_{\theta
ightarrow0}\frac{3\theta+cos\theta}{5sin\theta + 1}) d. (lim_{x
ightarrowinfty}(\frac{-3}{x^{4}}+1)) e. (lim_{x
ightarrow5^{+}}\frac{-1}{(x - 5)^{2}}) f. (lim_{x
ightarrow5^{+}}\frac{-1}{x - 5}) g. (lim_{x
ightarrow5^{+}}\frac{1}{x - 5}) h. (lim_{x
ightarrow5^{-}}\frac{1}{x - 5}) i. (lim_{\theta
ightarrow\frac{pi}{2}}\frac{cos\theta}{cot\theta}) 3. which limit(s) in question #2, are finding the horizontal asymptotes of the function? 4. which limit(s) in question #2, are finding the vertical asymptotes of the function? 5. find the limit. you must show how the property (lim_{\theta
ightarrow0}\frac{sin\theta}{\theta}=1) is used to find the limit. no shortcuts will earn points! a. (lim_{\theta
ightarrow0}\frac{sin(3\theta)}{sin(5\theta)}) b. (lim_{\theta
ightarrow0}\frac{\tan(5\theta)}{3\theta}) c. (lim_{\theta
ightarrow0}\frac{sin(3\theta)}{2\theta})

Explanation:

Step1: Solve 2a

Substitute \(x = - 1\) into \(2x^{2}-3x\).
\[2(-1)^{2}-3(-1)=2\times1 + 3=5\]

Step2: Solve 2b

Factor the numerator and denominator. \(x^{2}-2x - 3=(x - 3)(x+1)\) and \(x^{2}-5x + 6=(x - 3)(x - 2)\). Then \(\lim_{x
ightarrow3}\frac{x^{2}-2x - 3}{x^{2}-5x + 6}=\lim_{x
ightarrow3}\frac{(x - 3)(x + 1)}{(x - 3)(x - 2)}=\lim_{x
ightarrow3}\frac{x + 1}{x - 2}=\frac{3+1}{3 - 2}=4\)

Step3: Solve 2c

Substitute \(\theta=0\) into \(\frac{3\theta+\cos\theta}{5\sin\theta + 1}\).
\(\frac{3\times0+\cos0}{5\sin0+1}=\frac{0 + 1}{0+1}=1\)

Step4: Solve 2d

As \(x
ightarrow\infty\), \(\lim_{x
ightarrow\infty}\frac{-3}{x^{4}}=0\), so \(\lim_{x
ightarrow\infty}(\frac{-3}{x^{4}}+1)=0 + 1=1\)

Step5: Solve 2e

As \(x
ightarrow5^{+}\), \((x - 5)^{2}
ightarrow0^{+}\), so \(\lim_{x
ightarrow5^{+}}\frac{-1}{(x - 5)^{2}}=-\infty\)

Step6: Solve 2f

As \(x
ightarrow5^{+}\), \(x-5
ightarrow0^{+}\), so \(\lim_{x
ightarrow5^{+}}\frac{-1}{x - 5}=-\infty\)

Step7: Solve 2g

As \(x
ightarrow5^{+}\), \(x - 5
ightarrow0^{+}\), so \(\lim_{x
ightarrow5^{+}}\frac{1}{x - 5}=\infty\)

Step8: Solve 2h

As \(x
ightarrow5^{-}\), \(x - 5
ightarrow0^{-}\), so \(\lim_{x
ightarrow5^{-}}\frac{1}{x - 5}=-\infty\)

Step9: Solve 2i

Rewrite \(\frac{\cos\theta}{\cot\theta}=\frac{\cos\theta}{\frac{\cos\theta}{\sin\theta}}=\sin\theta\). Then \(\lim_{\theta
ightarrow\frac{\pi}{2}}\frac{\cos\theta}{\cot\theta}=\lim_{\theta
ightarrow\frac{\pi}{2}}\sin\theta = 1\)

Step10: Solve 5a

Use the formula \(\lim_{\theta
ightarrow0}\frac{\sin\alpha\theta}{\alpha\theta}=1\). \(\lim_{\theta
ightarrow0}\frac{\sin(3\theta)}{\sin(5\theta)}=\lim_{\theta
ightarrow0}\frac{\sin(3\theta)}{3\theta}\times\frac{5\theta}{\sin(5\theta)}\times\frac{3}{5}=\frac{3}{5}\)

Step11: Solve 5b

Since \(\tan(5\theta)=\frac{\sin(5\theta)}{\cos(5\theta)}\), \(\lim_{\theta
ightarrow0}\frac{\tan(5\theta)}{3\theta}=\lim_{\theta
ightarrow0}\frac{\sin(5\theta)}{3\theta\cos(5\theta)}=\lim_{\theta
ightarrow0}\frac{\sin(5\theta)}{5\theta}\times\frac{5}{3}\times\frac{1}{\cos(5\theta)}=\frac{5}{3}\)

Step12: Solve 5c

\(\lim_{\theta
ightarrow0}\frac{\sin(3\theta)}{2\theta}=\lim_{\theta
ightarrow0}\frac{\sin(3\theta)}{3\theta}\times\frac{3}{2}=\frac{3}{2}\)

Step13: Answer 3

The limits in 2d (\(\lim_{x
ightarrow\infty}(\frac{-3}{x^{4}}+1)\)) are finding the horizontal - asymptotes of the function. Because for a function \(y = f(x)\), \(\lim_{x
ightarrow\pm\infty}f(x)=L\) gives the horizontal asymptote \(y = L\).

Step14: Answer 4

The limits in 2e (\(\lim_{x
ightarrow5^{+}}\frac{-1}{(x - 5)^{2}}\)), 2f (\(\lim_{x
ightarrow5^{+}}\frac{-1}{x - 5}\)), 2g (\(\lim_{x
ightarrow5^{+}}\frac{1}{x - 5}\)) and 2h (\(\lim_{x
ightarrow5^{-}}\frac{1}{x - 5}\)) are finding the vertical asymptotes of the function. Because for a function \(y = f(x)\), if \(\lim_{x
ightarrow a^{+}}f(x)=\pm\infty\) or \(\lim_{x
ightarrow a^{-}}f(x)=\pm\infty\), then \(x = a\) is a vertical asymptote.

Answer:

2a. 5
2b. 4
2c. 1
2d. 1
2e. \(-\infty\)
2f. \(-\infty\)
2g. \(\infty\)
2h. \(-\infty\)
2i. 1
5a. \(\frac{3}{5}\)
5b. \(\frac{5}{3}\)
5c. \(\frac{3}{2}\)

1. 2d
2. 2e, 2f, 2g, 2h