evaluate each limit below. your answer should be a finite number, +∞, -∞, or does not exist.\n4…

evaluate each limit below. your answer should be a finite number, +∞, -∞, or does not exist.\n4. $lim_{x\rightarrow2^{+}}\frac{3x^{2}-6x}{|2 - x|}$\n5. $lim_{x\rightarrow0}\frac{e^{4x}-1}{sin4x}$\n6. $lim_{x\rightarrow3}\frac{2x - 6}{sqrt{x}-sqrt{3}}$\n7. $lim_{x\rightarrow1}\frac{x^{2}-3x + 2}{x^{2}+2x - 3}$\n8. $lim_{x\rightarrow1}(x - 1)^{4}sin(\frac{pi}{1 - x})$\n9. $lim_{x\rightarrow0}\frac{4x}{sin(2x)cos x}$\n10. $lim_{x\rightarrow1^{+}}\frac{x^{2}-3x + 2}{x^{2}-2x + 1}$\n11. $lim_{x\rightarrow0}(\frac{3}{x^{2}+3x}-\frac{1}{x})$\ncaution: 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.

evaluate each limit below. your answer should be a finite number, +∞, -∞, or does not exist.\n4. $lim_{x\rightarrow2^{+}}\frac{3x^{2}-6x}{|2 - x|}$\n5. $lim_{x\rightarrow0}\frac{e^{4x}-1}{sin4x}$\n6. $lim_{x\rightarrow3}\frac{2x - 6}{sqrt{x}-sqrt{3}}$\n7. $lim_{x\rightarrow1}\frac{x^{2}-3x + 2}{x^{2}+2x - 3}$\n8. $lim_{x\rightarrow1}(x - 1)^{4}sin(\frac{pi}{1 - x})$\n9. $lim_{x\rightarrow0}\frac{4x}{sin(2x)cos x}$\n10. $lim_{x\rightarrow1^{+}}\frac{x^{2}-3x + 2}{x^{2}-2x + 1}$\n11. $lim_{x\rightarrow0}(\frac{3}{x^{2}+3x}-\frac{1}{x})$\ncaution: 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.

Answer

4. $\lim_{x\rightarrow2^{+}}\frac{3x^{2}-6x}{|2 - x|}$

Explanation:

Step1: Analyze the absolute - value

When $x\rightarrow2^{+}$, $2 - x<0$, so $|2 - x|=-(2 - x)=x - 2$. Then the function becomes $\lim_{x\rightarrow2^{+}}\frac{3x(x - 2)}{x - 2}$.

Step2: Simplify the function

Cancel out the common factor $(x - 2)$ (since $x\neq2$ when taking the limit), we get $\lim_{x\rightarrow2^{+}}3x$.

Step3: Evaluate the limit

Substitute $x = 2$ into $3x$, we have $3\times2=6$.

5. $\lim_{x\rightarrow0}\frac{e^{4x}-1}{\sin4x}$

Explanation:

Step1: Use the well - known limit

We know that $\lim_{u\rightarrow0}\frac{e^{u}-1}{u}=1$ and $\lim_{u\rightarrow0}\frac{\sin u}{u}=1$. Let $u = 4x$. As $x\rightarrow0$, $u\rightarrow0$. Then $\lim_{x\rightarrow0}\frac{e^{4x}-1}{\sin4x}=\lim_{x\rightarrow0}\frac{e^{4x}-1}{4x}\times\frac{4x}{\sin4x}$.

Step2: Evaluate the product of limits

Since $\lim_{x\rightarrow0}\frac{e^{4x}-1}{4x}=1$ and $\lim_{x\rightarrow0}\frac{4x}{\sin4x}=1$, the limit is $1\times1 = 1$.

6. $\lim_{x\rightarrow3}\frac{2x - 6}{\sqrt{x}-\sqrt{3}}$

Explanation:

Step1: Rationalize the denominator

Multiply the numerator and denominator by $\sqrt{x}+\sqrt{3}$: $\lim_{x\rightarrow3}\frac{(2x - 6)(\sqrt{x}+\sqrt{3})}{x - 3}$.

Step2: Factor the numerator

Factor out 2 from the numerator: $\lim_{x\rightarrow3}\frac{2(x - 3)(\sqrt{x}+\sqrt{3})}{x - 3}$.

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. $\lim_{x\rightarrow1}\frac{x^{2}-3x + 2}{x^{2}+2x - 3}$

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 $x\neq1$ when taking the limit), we get $\lim_{x\rightarrow1}\frac{x - 2}{x + 3}$.

Step3: Evaluate the limit

Substitute $x = 1$ into $\frac{x - 2}{x + 3}$, we have $\frac{1-2}{1 + 3}=-\frac{1}{4}$.

8. $\lim_{x\rightarrow1}(x - 1)^{4}\sin(\frac{\pi}{1 - x})$

Explanation:

Step1: Let $t=1 - x$

As $x\rightarrow1$, $t\rightarrow0$. Then the limit becomes $\lim_{t\rightarrow0}t^{4}\sin(\frac{\pi}{t})$.

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 $\lim_{t\rightarrow0}-t^{4}=0$ and $\lim_{t\rightarrow0}t^{4}=0$, by the Squeeze Theorem, $\lim_{t\rightarrow0}t^{4}\sin(\frac{\pi}{t})=0$.

9. $\lim_{x\rightarrow0}\frac{4x}{\sin(2x)\cos x}$

Explanation:

Step1: Rewrite the limit

We know that $\sin(2x)=2\sin x\cos x$. So the limit becomes $\lim_{x\rightarrow0}\frac{4x}{2\sin x\cos x\cos x}=\lim_{x\rightarrow0}\frac{2x}{\sin x\cos^{2}x}$.

Step2: Split the limit

$\lim_{x\rightarrow0}\frac{2x}{\sin x\cos^{2}x}=\lim_{x\rightarrow0}\frac{2x}{\sin x}\times\lim_{x\rightarrow0}\frac{1}{\cos^{2}x}$.

Step3: Evaluate the limits

Since $\lim_{x\rightarrow0}\frac{x}{\sin x}=1$ and $\lim_{x\rightarrow0}\cos x = 1$, we have $2\times1\times1 = 2$.

10. $\lim_{x\rightarrow1^{+}}\frac{x^{2}-3x + 2}{x^{2}-2x + 1}$

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 $\lim_{x\rightarrow1^{+}}\frac{x - 2}{x - 1}$.

Step3: Evaluate the limit

As $x\rightarrow1^{+}$, the numerator approaches $1-2=-1$ and the denominator approaches $0^{+}$. So $\lim_{x\rightarrow1^{+}}\frac{x - 2}{x - 1}=-\infty$.

11. $\lim_{x\rightarrow0}(\frac{3}{x^{2}+3x}-\frac{1}{x})$

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 $x\neq0$ when taking the limit), we get $\lim_{x\rightarrow0}\frac{-1}{x + 3}$.

Step3: Evaluate the limit

Substitute $x = 0$ into $\frac{-1}{x + 3}$, we have $\frac{-1}{0+3}=-\frac{1}{3}$.

Answer:

  1. $6$
  2. $1$
  3. $4\sqrt{3}$
  4. $-\frac{1}{4}$
  5. $0$
  6. $2$
  7. $-\infty$
  8. $-\frac{1}{3}$