Question

i. find the limits:
(1) $lim_{x
ightarrow0}\frac{sin3x}{sin x}$
(2) $lim_{x
ightarrowinfty}\frac{x}{x + 1}$
(3) $lim_{x
ightarrow0}\frac{sin4x}{x^{2}}$
(4) $lim_{x
ightarrowinfty}\frac{x^{2}-1}{ln x}$
(5) $lim_{x
ightarrowinfty}\frac{x^{2}}{e^{x}}$
(6) $lim_{x
ightarrowinfty}\frac{x^{2}+2}{7x^{2}}$

ii. find each indefinite integral:
(1) $int(3 + x^{4})dx$
(2) $intsec x\tan xdx$
(3) $int(x+\frac{1}{x})dx$

Explanation:

I. Find the limits:

(1) $\lim_{x

ightarrow0}\frac{\sin3x}{\sin x}$

Step1: Use the double - angle formula $\sin A = A\cdot\frac{\sin A}{A}$

$\lim_{x
ightarrow0}\frac{\sin3x}{\sin x}=\lim_{x
ightarrow0}\frac{3x\cdot\frac{\sin3x}{3x}}{x\cdot\frac{\sin x}{x}}$

Step2: Evaluate the limits of $\frac{\sin3x}{3x}$ and $\frac{\sin x}{x}$

Since $\lim_{u
ightarrow0}\frac{\sin u}{u} = 1$, when $u = 3x$ and $u=x$ as $x
ightarrow0$.
$\lim_{x
ightarrow0}\frac{3x\cdot\frac{\sin3x}{3x}}{x\cdot\frac{\sin x}{x}}=\lim_{x
ightarrow0}\frac{3\cdot\frac{\sin3x}{3x}}{\frac{\sin x}{x}}=3$

(2) $\lim_{x

ightarrow\infty}\frac{x}{x + 1}$

Step1: Divide numerator and denominator by $x$

$\lim_{x
ightarrow\infty}\frac{x}{x + 1}=\lim_{x
ightarrow\infty}\frac{\frac{x}{x}}{\frac{x}{x}+\frac{1}{x}}=\lim_{x
ightarrow\infty}\frac{1}{1+\frac{1}{x}}$

Step2: Evaluate the limit of $\frac{1}{x}$ as $x

ightarrow\infty$
Since $\lim_{x
ightarrow\infty}\frac{1}{x}=0$, then $\lim_{x
ightarrow\infty}\frac{1}{1+\frac{1}{x}} = 1$

(3) $\lim_{x

ightarrow0}\frac{\sin4x}{x^{2}}$

Step1: Use $\sin4x=4x\cdot\frac{\sin4x}{4x}$

$\lim_{x
ightarrow0}\frac{\sin4x}{x^{2}}=\lim_{x
ightarrow0}\frac{4x\cdot\frac{\sin4x}{4x}}{x^{2}}=\lim_{x
ightarrow0}\frac{4\cdot\frac{\sin4x}{4x}}{x}$

Step2: Analyze the limit

As $x
ightarrow0$, $\frac{\sin4x}{4x}
ightarrow1$, and $\lim_{x
ightarrow0}\frac{4\cdot\frac{\sin4x}{4x}}{x}=\infty$

(4) $\lim_{x

ightarrow+\infty}\frac{x^{2}-1}{\ln x}$

Step1: Apply L'Hopital's rule (since it is in $\frac{\infty}{\infty}$ form)

Differentiate the numerator and denominator. The derivative of $y = x^{2}-1$ is $y^\prime=2x$, and the derivative of $y=\ln x$ is $y^\prime=\frac{1}{x}$
$\lim_{x
ightarrow+\infty}\frac{x^{2}-1}{\ln x}=\lim_{x
ightarrow+\infty}\frac{2x}{\frac{1}{x}}=\lim_{x
ightarrow+\infty}2x^{2}=\infty$

(5) $\lim_{x

ightarrow+\infty}\frac{x^{2}}{e^{x}}$

Step1: Apply L'Hopital's rule (since it is in $\frac{\infty}{\infty}$ form)

Differentiate the numerator and denominator. The derivative of $y = x^{2}$ is $y^\prime = 2x$, and the derivative of $y=e^{x}$ is $y^\prime=e^{x}$
$\lim_{x
ightarrow+\infty}\frac{x^{2}}{e^{x}}=\lim_{x
ightarrow+\infty}\frac{2x}{e^{x}}$

Step2: Apply L'Hopital's rule again

Differentiate the new numerator and denominator. The derivative of $y = 2x$ is $y^\prime=2$, and the derivative of $y = e^{x}$ is $y^\prime=e^{x}$
$\lim_{x
ightarrow+\infty}\frac{2x}{e^{x}}=\lim_{x
ightarrow+\infty}\frac{2}{e^{x}}=0$

(6) $\lim_{x

ightarrow+\infty}\frac{x^{2}+2}{7x^{2}}$

Step1: Divide numerator and denominator by $x^{2}$

$\lim_{x
ightarrow+\infty}\frac{x^{2}+2}{7x^{2}}=\lim_{x
ightarrow+\infty}\frac{1+\frac{2}{x^{2}}}{7}$

Step2: Evaluate the limit of $\frac{2}{x^{2}}$ as $x

ightarrow\infty$
Since $\lim_{x
ightarrow\infty}\frac{2}{x^{2}} = 0$, then $\lim_{x
ightarrow+\infty}\frac{1+\frac{2}{x^{2}}}{7}=\frac{1}{7}$

II. Find each indefinite integral:

(1) $\int(3 + x^{4})dx$

Step1: Use the sum - rule of integration $\int(f(x)+g(x))dx=\int f(x)dx+\int g(x)dx$

$\int(3 + x^{4})dx=\int3dx+\int x^{4}dx$

Step2: Apply the power - rule of integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n

eq - 1)$ and $\int kdx=kx + C$ ($k$ is a constant)
$\int3dx=3x+C_1$, $\int x^{4}dx=\frac{x^{5}}{5}+C_2$
So $\int(3 + x^{4})dx=3x+\frac{x^{5}}{5}+C$ ($C = C_1 + C_2$)

(2) $\int\sec x\tan xdx$

Step1: Recall the derivative formula

We know that the derivative of $\sec x$ is $\sec x\tan x$. So $\int\sec x\tan xdx=\sec x + C$

(3) $\int(x+\frac{1}{x})dx$

Step1: Use the sum - rule of integration $\int(f(x)+g(x))dx=\int f(x)dx+\int g(x)dx$

$\int(x+\frac{1}{x})dx=\int xdx+\int\frac{1}{x}dx$

Step2: Apply the power - rule of integration for $\int xdx$ and the formula for $\int\frac{1}{x}dx$

$\int xdx=\frac{x^{2}}{2}+C_1$, $\int\frac{1}{x}dx=\ln|x|+C_2$
So $\int(x+\frac{1}{x})dx=\frac{x^{2}}{2}+\ln|x|+C$ ($C = C_1 + C_2$)

Answer:

I.

(1) $3$
(2) $1$
(3) $\infty$
(4) $\infty$
(5) $0$
(6) $\frac{1}{7}$

II.

(1) $3x+\frac{x^{5}}{5}+C$
(2) $\sec x + C$
(3) $\frac{x^{2}}{2}+\ln|x|+C$