find the following limits (enter the answers in the blank fields; write dne if a limit does not exist):\n1…

Explanation:

1. $\lim_{x\rightarrow2}\frac{x^{2}+x - 6}{x - 2}$

Step1: Factor the numerator

$x^{2}+x - 6=(x + 3)(x - 2)$ So, $\frac{x^{2}+x - 6}{x - 2}=\frac{(x + 3)(x - 2)}{x - 2}=x + 3$ (for $x\neq2$)

Step2: Evaluate the limit

$\lim_{x\rightarrow2}(x + 3)=2+3 = 5$

2. $\lim_{x\rightarrow0}\frac{4e^{x}-4 - 4x}{x^{2}}$

Step1: Apply L'Hopital's Rule (since $\frac{0}{0}$ form)

Differentiate numerator and denominator. The derivative of $4e^{x}-4 - 4x$ is $4e^{x}-4$, and the derivative of $x^{2}$ is $2x$. So, $\lim_{x\rightarrow0}\frac{4e^{x}-4}{2x}$ (still $\frac{0}{0}$ form)

Step2: Apply L'Hopital's Rule again

Differentiate numerator and denominator. The derivative of $4e^{x}-4$ is $4e^{x}$, and the derivative of $2x$ is $2$. Then $\lim_{x\rightarrow0}\frac{4e^{x}}{2}=\frac{4e^{0}}{2}=2$

3. $\lim_{x\rightarrow0}\frac{\arcsin(9x)}{\tan(3x)}$

Step1: Use equivalent - infinitesimals

As $u\rightarrow0$, $\arcsin(u)\sim u$ and $\tan(u)\sim u$. When $x\rightarrow0$, $\arcsin(9x)\sim9x$ and $\tan(3x)\sim3x$

Step2: Evaluate the limit

$\lim_{x\rightarrow0}\frac{\arcsin(9x)}{\tan(3x)}=\lim_{x\rightarrow0}\frac{9x}{3x}=3$

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

Step1: Substitute $x = 1$

$x^{2}-1=(x + 1)(x - 1)$, so $\frac{x^{2}-1}{\arctan x}=\frac{(x + 1)(x - 1)}{\arctan x}$ Substitute $x = 1$: $\frac{(1 + 1)(1 - 1)}{\arctan(1)}=\frac{0}{\frac{\pi}{4}}=0$

5. $\lim_{x\rightarrow1}\pi\ln(x)\csc(\pi x)$

Step1: Rewrite $\csc(\pi x)=\frac{1}{\sin(\pi x)}$

Let $t=x - 1$, then $x=t + 1$. As $x\rightarrow1$, $t\rightarrow0$ $\ln(x)=\ln(1 + t)\sim t$ and $\sin(\pi x)=\sin(\pi(t + 1))=-\sin(\pi t)\sim-\pi t$

Step2: Evaluate the limit

$\lim_{x\rightarrow1}\pi\ln(x)\csc(\pi x)=\lim_{t\rightarrow0}\pi\frac{\ln(1 + t)}{\sin(\pi(1 + t))}=\lim_{t\rightarrow0}\pi\frac{t}{-\pi t}=- 1$

6. $\lim_{x\rightarrow0}\cot(3x)\sin(12x)$

Step1: Rewrite $\cot(3x)=\frac{\cos(3x)}{\sin(3x)}$

$\cot(3x)\sin(12x)=\frac{\cos(3x)\sin(12x)}{\sin(3x)}$

Step2: Use $\sin(A)=2\sin\frac{A}{2}\cos\frac{A}{2}$ and equivalent - infinitesimals

$\sin(12x)=2\sin(6x)\cos(6x)=4\sin(3x)\cos(3x)\cos(6x)$ (using $\sin(2\alpha)=2\sin\alpha\cos\alpha$) $\lim_{x\rightarrow0}\frac{\cos(3x)\sin(12x)}{\sin(3x)}=\lim_{x\rightarrow0}\frac{\cos(3x)\times4\sin(3x)\cos(3x)\cos(6x)}{\sin(3x)} = 4$

7. $\lim_{x\rightarrow\infty}(\sqrt{x^{2}+8x}-x)$

Step1: Rationalize the expression

Multiply by $\frac{\sqrt{x^{2}+8x}+x}{\sqrt{x^{2}+8x}+x}$ $\frac{(\sqrt{x^{2}+8x}-x)(\sqrt{x^{2}+8x}+x)}{\sqrt{x^{2}+8x}+x}=\frac{x^{2}+8x-x^{2}}{\sqrt{x^{2}+8x}+x}=\frac{8x}{\sqrt{x^{2}+8x}+x}$

Step2: Divide numerator and denominator by $x$

$\frac{8}{\sqrt{1+\frac{8}{x}}+1}$

Step3: Evaluate the limit

$\lim_{x\rightarrow\infty}\frac{8}{\sqrt{1+\frac{8}{x}}+1}=\frac{8}{1 + 1}=4$

8. $\lim_{x\rightarrow\infty}(xe^{\frac{1}{x}}-x)$

Step1: Let $t=\frac{1}{x}$, then $x\rightarrow\infty$ implies $t\rightarrow0^{+}$

The limit becomes $\lim_{t\rightarrow0^{+}}\frac{e^{t}-1}{t}$

Step2: Use the fact that $\lim_{u\rightarrow0}\frac{e^{u}-1}{u}=1$ (by L'Hopital's Rule: $\lim_{u\rightarrow0}\frac{e^{u}-1}{u}=\lim_{u\rightarrow0}\frac{e^{u}}{1}=1$)

9. $\lim_{x\rightarrow\infty}x^{\frac{\ln(6)}{3+\ln(x)}}$

Step1: Take the natural logarithm

Let $y = x^{\frac{\ln(6)}{3+\ln(x)}}$, then $\ln y=\frac{\ln(6)\ln x}{3+\ln x}$

Step2: Divide numerator and denominator by $\ln x$

$\ln y=\frac{\ln(6)}{\frac{3}{\ln x}+1}$

Step3: Evaluate the limit of $\ln y$

$\lim_{x\rightarrow\infty}\ln y=\ln(6)$

Step4: Evaluate the limit of $y$

$\lim_{x\rightarrow\infty}y = e^{\ln(6)}=6$

10. $\lim_{x\rightarrow\infty}(1+\frac{\ln(7)}{x}+\frac{9}{x^{2}})^{x}$

Step1: Use the formula $\lim_{x\rightarrow\infty}(1+\frac{a}{x}+\frac{b}{x^{2}})^{x}=e^{a}$

Here $a = \ln(7)$ and $b = 9$ Since $\lim_{x\rightarrow\infty}(1+\frac{\ln(7)}{x}+\frac{9}{x^{2}})^{x}=\lim_{x\rightarrow\infty}\left[(1+\frac{\ln(7)+\frac{9}{x}}{x})^{x}\right]$ and $\lim_{x\rightarrow\infty}\frac{9}{x}=0$ By the formula $\lim_{t\rightarrow\infty}(1+\frac{c}{t})^{t}=e^{c}$, we have $\lim_{x\rightarrow\infty}(1+\frac{\ln(7)}{x}+\frac{9}{x^{2}})^{x}=7$

Answer:

1. $5$
2. $2$
3. $3$
4. $0$
5. $-1$
6. $4$
7. $4$
8. $1$
9. $6$
10. $7$