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

1. a. (y = e^{x - 3}); over (3,7) b. (y=sqrt{6 - 3x}); over (0,2) 2. find the limit. a. (lim_{x\rightarrow - 1}(2x^{2}-3x)) b. (lim_{x\rightarrow3}\frac{x^{2}-2x - 3}{x^{2}-5x + 6}) c. (lim_{\theta\rightarrow0}\frac{3\theta+cos\theta}{5sin\theta + 1}) d. (lim_{x\rightarrowinfty}(\frac{-3}{x^{4}}+1)) e. (lim_{x\rightarrow5^{+}}\frac{-1}{(x - 5)^{2}}) f. (lim_{x\rightarrow5^{+}}\frac{-1}{x - 5}) g. (lim_{x\rightarrow5^{+}}\frac{1}{x - 5}) h. (lim_{x\rightarrow5^{-}}\frac{1}{x - 5}) i. (lim_{\theta\rightarrow\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\rightarrow0}\frac{sin\theta}{\theta}=1) is used to find the limit. no shortcuts will earn points! a. (lim_{\theta\rightarrow0}\frac{sin(3\theta)}{sin(5\theta)}) b. (lim_{\theta\rightarrow0}\frac{\tan(5\theta)}{3\theta}) c. (lim_{\theta\rightarrow0}\frac{sin(3\theta)}{2\theta})

1. a. (y = e^{x - 3}); over (3,7) b. (y=sqrt{6 - 3x}); over (0,2) 2. find the limit. a. (lim_{x\rightarrow - 1}(2x^{2}-3x)) b. (lim_{x\rightarrow3}\frac{x^{2}-2x - 3}{x^{2}-5x + 6}) c. (lim_{\theta\rightarrow0}\frac{3\theta+cos\theta}{5sin\theta + 1}) d. (lim_{x\rightarrowinfty}(\frac{-3}{x^{4}}+1)) e. (lim_{x\rightarrow5^{+}}\frac{-1}{(x - 5)^{2}}) f. (lim_{x\rightarrow5^{+}}\frac{-1}{x - 5}) g. (lim_{x\rightarrow5^{+}}\frac{1}{x - 5}) h. (lim_{x\rightarrow5^{-}}\frac{1}{x - 5}) i. (lim_{\theta\rightarrow\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\rightarrow0}\frac{sin\theta}{\theta}=1) is used to find the limit. no shortcuts will earn points! a. (lim_{\theta\rightarrow0}\frac{sin(3\theta)}{sin(5\theta)}) b. (lim_{\theta\rightarrow0}\frac{\tan(5\theta)}{3\theta}) c. (lim_{\theta\rightarrow0}\frac{sin(3\theta)}{2\theta})

Answer

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\rightarrow3}\frac{x^{2}-2x - 3}{x^{2}-5x + 6}=\lim_{x\rightarrow3}\frac{(x - 3)(x + 1)}{(x - 3)(x - 2)}=\lim_{x\rightarrow3}\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\rightarrow\infty), (\lim_{x\rightarrow\infty}\frac{-3}{x^{4}}=0), so (\lim_{x\rightarrow\infty}(\frac{-3}{x^{4}}+1)=0 + 1=1)

Step5: Solve 2e

As (x\rightarrow5^{+}), ((x - 5)^{2}\rightarrow0^{+}), so (\lim_{x\rightarrow5^{+}}\frac{-1}{(x - 5)^{2}}=-\infty)

Step6: Solve 2f

As (x\rightarrow5^{+}), (x-5\rightarrow0^{+}), so (\lim_{x\rightarrow5^{+}}\frac{-1}{x - 5}=-\infty)

Step7: Solve 2g

As (x\rightarrow5^{+}), (x - 5\rightarrow0^{+}), so (\lim_{x\rightarrow5^{+}}\frac{1}{x - 5}=\infty)

Step8: Solve 2h

As (x\rightarrow5^{-}), (x - 5\rightarrow0^{-}), so (\lim_{x\rightarrow5^{-}}\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\rightarrow\frac{\pi}{2}}\frac{\cos\theta}{\cot\theta}=\lim_{\theta\rightarrow\frac{\pi}{2}}\sin\theta = 1)

Step10: Solve 5a

Use the formula (\lim_{\theta\rightarrow0}\frac{\sin\alpha\theta}{\alpha\theta}=1). (\lim_{\theta\rightarrow0}\frac{\sin(3\theta)}{\sin(5\theta)}=\lim_{\theta\rightarrow0}\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\rightarrow0}\frac{\tan(5\theta)}{3\theta}=\lim_{\theta\rightarrow0}\frac{\sin(5\theta)}{3\theta\cos(5\theta)}=\lim_{\theta\rightarrow0}\frac{\sin(5\theta)}{5\theta}\times\frac{5}{3}\times\frac{1}{\cos(5\theta)}=\frac{5}{3})

Step12: Solve 5c

(\lim_{\theta\rightarrow0}\frac{\sin(3\theta)}{2\theta}=\lim_{\theta\rightarrow0}\frac{\sin(3\theta)}{3\theta}\times\frac{3}{2}=\frac{3}{2})

Step13: Answer 3

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

Step14: Answer 4

The limits in 2e ((\lim_{x\rightarrow5^{+}}\frac{-1}{(x - 5)^{2}})), 2f ((\lim_{x\rightarrow5^{+}}\frac{-1}{x - 5})), 2g ((\lim_{x\rightarrow5^{+}}\frac{1}{x - 5})) and 2h ((\lim_{x\rightarrow5^{-}}\frac{1}{x - 5})) are finding the vertical asymptotes of the function. Because for a function (y = f(x)), if (\lim_{x\rightarrow a^{+}}f(x)=\pm\infty) or (\lim_{x\rightarrow 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}) 3. 2d 4. 2e, 2f, 2g, 2h