Evaluate the following limits. If the limit does not exist, explain why. $$ \begin{array}{lll} ext { (a) } \lim _{t ightarrow-2}\left(\frac{2-|t|}{2+t} ight) & ext { (b) } \lim _{x ightarrow 2^{-}} \frac{6-3 x}{|x-2|} & ext { (c) } \lim _{x ightarrow 3} \frac{1}{|3-x|} \ ext { (c) } \lim _{x ightarrow 0}\left(x^{2} \cos \left(\frac{2}{x}+4 ight) ight) & ext { (d) } \lim _{x ightarrow 0} x^{2}\left(1+\sin \frac{\pi}{x} ight) & ext { (e) } \lim _{x ightarrow 2^{+}} \sqrt{x-2} \cos \left(\frac{1}{x-2} ight) \ ext { (f) } \lim _{x ightarrow \infty} \sin \left(2 an ^{-1}(x) ight) & ext { (g) } \lim _{x ightarrow \infty} \arctan e^{x} & ext { (h) } \lim _{x ightarrow \infty} \arctan e^{-x}\end{array} $$

Question

Evaluate the following limits. If the limit does not exist, explain why. $$ \begin{array}{lll}	ext { (a) } \lim _{t ightarrow-2}\left(\frac{2-|t|}{2+t}ight) & 	ext { (b) } \lim _{x ightarrow 2^{-}} \frac{6-3 x}{|x-2|} & 	ext { (c) } \lim _{x ightarrow 3} \frac{1}{|3-x|} \ 	ext { (c) } \lim _{x ightarrow 0}\left(x^{2} \cos \left(\frac{2}{x}+4ight)ight) & 	ext { (d) } \lim _{x ightarrow 0} x^{2}\left(1+\sin \frac{\pi}{x}ight) & 	ext { (e) } \lim _{x ightarrow 2^{+}} \sqrt{x-2} \cos \left(\frac{1}{x-2}ight) \ 	ext { (f) } \lim _{x ightarrow \infty} \sin \left(2 	an ^{-1}(x)ight) & 	ext { (g) } \lim _{x ightarrow \infty} \arctan e^{x} & 	ext { (h) } \lim _{x ightarrow \infty} \arctan e^{-x}\end{array} $$

UpStudy AI · Accepted Answer

Evaluate the limit by following steps:
- step0: Evaluate the limit:
  $$\lim _{x\rightarrow +\infty}\left(\arctan\left(e^{-x}\right)\right)$$
- step1: Rewrite the expression:
  $$\arctan\left(\lim _{x\rightarrow +\infty}\left(e^{-x}\right)\right)$$
- step2: Calculate:
  $$\arctan\left(0\right)$$
- step3: Calculate the trigonometric value:
  $$0$$
Calculate the limit $$ \lim_{x \rightarrow 2^{+}} \sqrt{x - 2} \cos \left(\frac{1}{x - 2}\right) $$.
Evaluate the limit by following steps:
- step0: Evaluate the limit:
  $$\lim _{x\rightarrow 2^{+}}\left(\sqrt{x-2}\times \cos\left(\frac{1}{x-2}\right)\right)$$
- step1: Calculate:
  $$0$$
Calculate the limit $$ \lim_{x \rightarrow 0}\left(x^{2} \cos \left(\frac{2}{x} + 4\right)\right) $$.
Evaluate the limit by following steps:
- step0: Evaluate the limit:
  $$\lim _{x\rightarrow 0}\left(x^{2}\cos\left(\frac{2}{x}+4\right)\right)$$
- step1: Calculate:
  $$0$$
Calculate the limit $$ \lim_{t \rightarrow -2}\left(\frac{2 - |t|}{2 + t}\right) $$.
Evaluate the limit by following steps:
- step0: Evaluate the limit:
  $$\lim _{t\rightarrow -2}\left(\frac{2-\left|t\right|}{2+t}\right)$$
- step1: Evaluate the left-hand and the right-hand limits:
  $$\begin{align}&\lim _{t\rightarrow -2^{-}}\left(\frac{2-\left|t\right|}{2+t}\right)\&\lim _{t\rightarrow -2^{+}}\left(\frac{2-\left|t\right|}{2+t}\right)\end{align}$$
- step2: Evaluate the left-hand limit:
  $$\begin{align}&1\&1\end{align}$$
- step3: Calculate:
  $$1$$
Calculate the limit $$ \lim_{x \rightarrow 2^{-}} \frac{6 - 3x}{|x - 2|} $$.
Evaluate the limit by following steps:
- step0: Evaluate the limit:
  $$\lim _{x\rightarrow 2^{-}}\left(\frac{6-3x}{\left|x-2\right|}\right)$$
- step1: Rewrite the expression:
  $$\lim _{x\rightarrow 2^{-}}\left(\frac{6-3x}{-x+2}\right)$$
- step2: Reduce the fraction:
  $$\lim _{x\rightarrow 2^{-}}\left(3\right)$$
- step3: Calculate:
  $$3$$
Calculate the limit $$ \lim_{x \rightarrow \infty} \sin \left(2 \tan^{-1}(x)\right) $$.
Evaluate the limit by following steps:
- step0: Evaluate using transformations:
  $$\lim _{x\rightarrow +\infty}\left(\sin\left(2\arctan\left(x\right)\right)\right)$$
- step1: Rewrite the expression:
  $$\lim _{x\rightarrow +\infty}\left(2\sin\left(\arctan\left(x\right)\right)\cos\left(\arctan\left(x\right)\right)\right)$$
- step2: Rewrite the expression:
  $$\lim _{x\rightarrow +\infty}\left(2\sin\left(\arctan\left(x\right)\right)\times \frac{1}{\sqrt{1+x^{2}}}\right)$$
- step3: Rewrite the expression:
  $$\lim _{x\rightarrow +\infty}\left(2\times \frac{x}{\sqrt{1+x^{2}}}\times \frac{1}{\sqrt{1+x^{2}}}\right)$$
- step4: Multiply the terms:
  $$\lim _{x\rightarrow +\infty}\left(\frac{2x}{1+x^{2}}\right)$$
- step5: Rewrite the expression:
  $$\lim _{x\rightarrow +\infty}\left(\frac{x}{x^{2}}\times \frac{2}{\frac{1}{x^{2}}+1}\right)$$
- step6: Calculate:
  $$\lim _{x\rightarrow +\infty}\left(\frac{1}{x}\times \frac{2}{\frac{1}{x^{2}}+1}\right)$$
- step7: Rewrite the expression:
  $$\lim _{x\rightarrow +\infty}\left(\frac{1}{x}\right)\times \lim _{x\rightarrow +\infty}\left(\frac{2}{\frac{1}{x^{2}}+1}\right)$$
- step8: Calculate:
  $$\lim _{x\rightarrow +\infty}\left(\frac{1}{x}\right)\times \frac{\lim _{x\rightarrow +\infty}\left(2\right)}{\lim _{x\rightarrow +\infty}\left(\frac{1}{x^{2}}+1\right)}$$
- step9: Rewrite the expression:
  $$\lim _{x\rightarrow +\infty}\left(\frac{1}{x}\right)\times \frac{2}{\lim _{x\rightarrow +\infty}\left(\frac{1}{x^{2}}+1\right)}$$
- step10: Rewrite the expression:
  $$\lim _{x\rightarrow +\infty}\left(\frac{1}{x}\right)\times \frac{2}{1}$$
- step11: Rewrite the expression:
  $$0\times \frac{2}{1}$$
- step12: Calculate:
  $$0\times 2$$
- step13: Calculate:
  $$0$$
Calculate the limit $$ \lim_{x \rightarrow 0} x^{2}\left(1 + \sin \frac{\pi}{x}\right) $$.
Evaluate the limit by following steps:
- step0: Evaluate the limit:
  $$\lim _{x\rightarrow 0}\left(x^{2}\left(1+\sin\left(\frac{\pi }{x}\right)\right)\right)$$
- step1: Calculate:
  $$0$$
Calculate the limit $$ \lim_{x \rightarrow 3} \frac{1}{|3 - x|} $$.
Evaluate the limit by following steps:
- step0: Evaluate the limit:
  $$\lim _{x\rightarrow 3}\left(\frac{1}{\left|3-x\right|}\right)$$
- step1: Evaluate the left-hand and the right-hand limits:
  $$\begin{align}&\lim _{x\rightarrow 3^{-}}\left(\frac{1}{\left|3-x\right|}\right)\&\lim _{x\rightarrow 3^{+}}\left(\frac{1}{\left|3-x\right|}\right)\end{align}$$
- step2: Evaluate the left-hand limit:
  $$\begin{align}&+\infty\&+\infty\end{align}$$
- step3: Calculate:
  $$+\infty$$
Calculate the limit $$ \lim_{x \rightarrow \infty} \arctan e^{x} $$.
Evaluate the limit by following steps:
- step0: Evaluate the limit:
  $$\lim _{x\rightarrow +\infty}\left(\arctan\left(e^{x}\right)\right)$$
- step1: Rewrite the expression:
  $$\arctan\left(\lim _{x\rightarrow +\infty}\left(e^{x}\right)\right)$$
- step2: Calculate:
  $$\arctan\left(+\infty\right)$$
- step3: Calculate the trigonometric value:
  $$\frac{\pi }{2}$$
Let's evaluate each limit step by step and explain the results:

### (a) $$ \lim _{t \rightarrow -2}\left(\frac{2-|t|}{2+t}\right) $$

1. Substitute $$ t = -2 $$:
   $$
   \frac{2 - |-2|}{2 + (-2)} = \frac{2 - 2}{0} = \frac{0}{0}
   $$
   This is an indeterminate form.

2. Simplifying:
   $$
   |t| = -t \text{ for } t < 0 \Rightarrow \frac{2 - (-t)}{2 + t} = \frac{2 + t}{2 + t} = 1 \text{ for } t \neq -2
   $$

3. Therefore, the limit is:
   $$
   \lim_{t \rightarrow -2} 1 = 1
   $$

### (b) $$ \lim _{x \rightarrow 2^{-}} \frac{6-3x}{|x-2|} $$

1. As $$ x $$ approaches $$ 2 $$ from the left, $$ |x - 2| = 2 - x $$:
   $$
   \lim_{x \rightarrow 2^{-}} \frac{6 - 3x}{2 - x}
   $$

2. Substitute $$ x = 2 $$:
   $$
   \frac{6 - 3(2)}{2 - 2} = \frac{0}{0}
   $$
   This is an indeterminate form.

3. Simplifying:
   $$
   \frac{6 - 3x}{2 - x} = \frac{-3(x - 2)}{-(x - 2)} = 3 \text{ for } x \neq 2
   $$

4. Therefore, the limit is:
   $$
   \lim_{x \rightarrow 2^{-}} 3 = 3
   $$

### (c) $$ \lim _{x \rightarrow 3} \frac{1}{|3-x|} $$

1. As $$ x $$ approaches $$ 3 $$:
   $$
   |3 - x| = 3 - x \text{ for } x < 3 \text{ and } x - 3 \text{ for } x > 3
   $$

2. The limit approaches:
   $$
   \lim_{x \rightarrow 3} \frac{1}{|3 - x|} = \lim_{x \rightarrow 3} \frac{1}{0} = +\infty
   $$

### (d) $$ \lim _{x \rightarrow 0}\left(x^{2} \cos \left(\frac{2}{x}+4\right)\right) $$

1. As $$ x $$ approaches $$ 0 $$:
   $$
   x^{2} \text{ approaches } 0 \text{ and } \cos \left(\frac{2}{x}+4\right) \text{ oscillates between } -1 \text{ and } 1.
   $$

2. Therefore, the limit is:
   $$
   0 \cdot \text{(bounded oscillation)} = 0
   $$

### (e) $$ \lim _{x \rightarrow 0} x^{2}\left(1+\sin \frac{\pi}{x}\right) $$

1. As $$ x $$ approaches $$ 0 $$:
   $$
   x^{2} \text{ approaches } 0 \text{ and } \sin \frac{\pi}{x} \text{ oscillates between } -1 \text{ and } 1.
   $$

2. Therefore, the limit is:
   $$
   0 \cdot \text{(bounded oscillation)} = 0
   $$

### (f) $$ \lim _{x \rightarrow 2^{+}} \sqrt{x-2} \cos \left(\frac{1}{x-2}\right) $$

1. As $$ x $$ approaches $$ 2 $$ from the right:
   $$
   \sqrt{x - 2} \text{ approaches } 0 \text{ and } \cos \left(\frac{1}{x - 2}\right) \text{ oscillates between } -1 \text{ and } 1.
   $$

2. Therefore, the limit is:
   $$
   0 \cdot \text{(bounded oscillation)} = 0
   $$

### (g) $$ \lim _{x \rightarrow \infty} \sin \left(2 \tan^{-1}(x)\right) $$

1. As $$ x $$ approaches $$ \infty $$:
   $$
   \tan^{-1}(x) \text{ approaches } \frac{\pi}{2} \Rightarrow 2 \tan^{-1}(x) \text{ approaches } \pi.
   $$

2. Therefore, the limit is:
   $$
   \sin(\pi) = 0
   $$

### (h) $$ \lim _{x \rightarrow \infty} \arctan e^{x} $$

1. As $$ x $$ approaches $$ \infty $$:
   $$
   e^{x} \text{ approaches } \infty \Rightarrow \arctan(e^{x}) \text{ approaches } \frac{\pi}{2}.
   $$

### (i) $$ \lim _{x \rightarrow \infty} \arctan e^{-x} $$

1. As $$ x $$ approaches $$ \infty $$:
   $$
   e^{-x} \text{ approaches } 0 \Rightarrow \arctan(e^{-x}) \text{ approaches } 0.
   $$

### Summary of Results:
- (a) $$ 1 $$
- (b) $$ 3 $$
- (c) $$ +\infty $$
- (d) $$ 0 $$
- (e) $$ 0 $$
- (f) $$ 0 $$
- (g) $$ 0 $$
- (h) $$ \frac{\pi}{2} $$
- (i) $$ 0 $$
Here are the limits evaluated: - (a) 1 - (b) 3 - (c) Infinity - (d) 0 - (e) 0 - (f) 0 - (g) 0 - (h) π⁄2 - (i) 0

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