evaluate each limit below. your answer should be a finite number, +∞, -∞, or does not exist.
4. $lim_{x
ightarrow2^{+}}\frac{3x^{2}-6x}{|2 - x|}$
5. $lim_{x
ightarrow0}\frac{e^{4x}-1}{sin4x}$
6. $lim_{x
ightarrow3}\frac{2x - 6}{sqrt{x}-sqrt{3}}$
7. $lim_{x
ightarrow1}\frac{x^{2}-3x + 2}{x^{2}+2x - 3}$
8. $lim_{x
ightarrow1}(x - 1)^{4}sin(\frac{pi}{1 - x})$
9. $lim_{x
ightarrow0}\frac{4x}{sin(2x)cos x}$
10. $lim_{x
ightarrow1^{+}}\frac{x^{2}-3x + 2}{x^{2}-2x + 1}$
11. $lim_{x
ightarrow0}(\frac{3}{x^{2}+3x}-\frac{1}{x})$
caution: this should not be your only study tool. this is a sample of questions for the exam and the actual exam will have different questions, so it’s a good idea to review a variety of materials. make sure to study your homework, quizzes, lecture notes, and other resources.

Question

Accepted Answer

### 4. $\lim_{x
ightarrow2^{+}}\frac{3x^{2}-6x}{|2 - x|}$
# Explanation:
## Step1: Analyze the absolute - value
When $x
ightarrow2^{+}$, $2 - x<0$, so $|2 - x|=-(2 - x)=x - 2$. Then the function becomes $\lim_{x
ightarrow2^{+}}\frac{3x(x - 2)}{x - 2}$.
## Step2: Simplify the function
Cancel out the common factor $(x - 2)$ (since $x
eq2$ when taking the limit), we get $\lim_{x
ightarrow2^{+}}3x$.
## Step3: Evaluate the limit
Substitute $x = 2$ into $3x$, we have $3	imes2=6$.

### 5. $\lim_{x
ightarrow0}\frac{e^{4x}-1}{\sin4x}$
# Explanation:
## Step1: Use the well - known limit
We know that $\lim_{u
ightarrow0}\frac{e^{u}-1}{u}=1$ and $\lim_{u
ightarrow0}\frac{\sin u}{u}=1$. Let $u = 4x$. As $x
ightarrow0$, $u
ightarrow0$. Then $\lim_{x
ightarrow0}\frac{e^{4x}-1}{\sin4x}=\lim_{x
ightarrow0}\frac{e^{4x}-1}{4x}	imes\frac{4x}{\sin4x}$.
## Step2: Evaluate the product of limits
Since $\lim_{x
ightarrow0}\frac{e^{4x}-1}{4x}=1$ and $\lim_{x
ightarrow0}\frac{4x}{\sin4x}=1$, the limit is $1	imes1 = 1$.

### 6. $\lim_{x
ightarrow3}\frac{2x - 6}{\sqrt{x}-\sqrt{3}}$
# Explanation:
## Step1: Rationalize the denominator
Multiply the numerator and denominator by $\sqrt{x}+\sqrt{3}$: $\lim_{x
ightarrow3}\frac{(2x - 6)(\sqrt{x}+\sqrt{3})}{x - 3}$.
## Step2: Factor the numerator
Factor out 2 from the numerator: $\lim_{x
ightarrow3}\frac{2(x - 3)(\sqrt{x}+\sqrt{3})}{x - 3}$.
## Step3: Simplify and evaluate the limit
Cancel out the common factor $(x - 3)$ and substitute $x = 3$ into $2(\sqrt{x}+\sqrt{3})$, we get $2(\sqrt{3}+\sqrt{3})=4\sqrt{3}$.

### 7. $\lim_{x
ightarrow1}\frac{x^{2}-3x + 2}{x^{2}+2x - 3}$
# Explanation:
## Step1: Factor the numerator and denominator
Factor the numerator $x^{2}-3x + 2=(x - 1)(x - 2)$ and the denominator $x^{2}+2x - 3=(x - 1)(x+3)$.
## Step2: Simplify the function
Cancel out the common factor $(x - 1)$ (since $x
eq1$ when taking the limit), we get $\lim_{x
ightarrow1}\frac{x - 2}{x + 3}$.
## Step3: Evaluate the limit
Substitute $x = 1$ into $\frac{x - 2}{x + 3}$, we have $\frac{1-2}{1 + 3}=-\frac{1}{4}$.

### 8. $\lim_{x
ightarrow1}(x - 1)^{4}\sin(\frac{\pi}{1 - x})$
# Explanation:
## Step1: Let $t=1 - x$
As $x
ightarrow1$, $t
ightarrow0$. Then the limit becomes $\lim_{t
ightarrow0}t^{4}\sin(\frac{\pi}{t})$.
## Step2: Use the property of sine function
We know that $-1\leqslant\sin(\frac{\pi}{t})\leqslant1$. Then $-t^{4}\leqslant t^{4}\sin(\frac{\pi}{t})\leqslant t^{4}$.
## Step3: Apply the Squeeze Theorem
Since $\lim_{t
ightarrow0}-t^{4}=0$ and $\lim_{t
ightarrow0}t^{4}=0$, by the Squeeze Theorem, $\lim_{t
ightarrow0}t^{4}\sin(\frac{\pi}{t})=0$.

### 9. $\lim_{x
ightarrow0}\frac{4x}{\sin(2x)\cos x}$
# Explanation:
## Step1: Rewrite the limit
We know that $\sin(2x)=2\sin x\cos x$. So the limit becomes $\lim_{x
ightarrow0}\frac{4x}{2\sin x\cos x\cos x}=\lim_{x
ightarrow0}\frac{2x}{\sin x\cos^{2}x}$.
## Step2: Split the limit
$\lim_{x
ightarrow0}\frac{2x}{\sin x\cos^{2}x}=\lim_{x
ightarrow0}\frac{2x}{\sin x}	imes\lim_{x
ightarrow0}\frac{1}{\cos^{2}x}$.
## Step3: Evaluate the limits
Since $\lim_{x
ightarrow0}\frac{x}{\sin x}=1$ and $\lim_{x
ightarrow0}\cos x = 1$, we have $2	imes1	imes1 = 2$.

### 10. $\lim_{x
ightarrow1^{+}}\frac{x^{2}-3x + 2}{x^{2}-2x + 1}$
# Explanation:
## Step1: Factor the numerator and denominator
Factor the numerator $x^{2}-3x + 2=(x - 1)(x - 2)$ and the denominator $x^{2}-2x + 1=(x - 1)^{2}$.
## Step2: Simplify the function
The function becomes $\lim_{x
ightarrow1^{+}}\frac{x - 2}{x - 1}$.
## Step3: Evaluate the limit
As $x
ightarrow1^{+}$, the numerator approaches $1-2=-1$ and the denominator approaches $0^{+}$. So $\lim_{x
ightarrow1^{+}}\frac{x - 2}{x - 1}=-\infty$.

### 11. $\lim_{x
ightarrow0}(\frac{3}{x^{2}+3x}-\frac{1}{x})$
# Explanation:
## Step1: Find a common denominator
$\frac{3}{x^{2}+3x}-\frac{1}{x}=\frac{3-(x + 3)}{x(x + 3)}=\frac{3-x - 3}{x(x + 3)}=\frac{-x}{x(x + 3)}$.
## Step2: Simplify the function
Cancel out the common factor $x$ (since $x
eq0$ when taking the limit), we get $\lim_{x
ightarrow0}\frac{-1}{x + 3}$.
## Step3: Evaluate the limit
Substitute $x = 0$ into $\frac{-1}{x + 3}$, we have $\frac{-1}{0+3}=-\frac{1}{3}$.

# Answer:
4. $6$
5. $1$
6. $4\sqrt{3}$
7. $-\frac{1}{4}$
8. $0$
9. $2$
10. $-\infty$
11. $-\frac{1}{3}$

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QUESTION IMAGE

Question

Explanation:

4. $\lim_{x

Step1: Analyze the absolute - value

Step2: Simplify the function

Step3: Evaluate the limit

5. $\lim_{x

Step1: Use the well - known limit

Step2: Evaluate the product of limits

6. $\lim_{x

Step1: Rationalize the denominator

Step2: Factor the numerator

Step3: Simplify and evaluate the limit

7. $\lim_{x

Answer: