# Find the limit: lim_{xrightarrowinfty}frac{3x^6-x^4+2x^2-3}{4x^5+2x^3-5}

Question
Limits and continuity
Find the limit:
$$\lim_{x\rightarrow\infty}\frac{3x^6-x^4+2x^2-3}{4x^5+2x^3-5}$$

2021-01-03
$$\lim_{x\rightarrow\infty}\frac{3x^6-x^4+2x^2-3}{4x^5+2x^3-5}$$
Multiply and divided by $$x^5$$:
$$\lim_{x\rightarrow\infty}\frac{3x^6-x^4+2x^2-3}{4x^5+2x^3-5}=\lim_{x\rightarrow\infty}\frac{x^5\frac{3x^6-x^4+2x^2-3}{x^5}}{x^5\frac{4x^5+2x^3-5}{x^5}}$$
Divide:
$$\lim_{x\rightarrow\infty}\frac{x^5\frac{3x^6-x^4+2x^2-3}{x^5}}{x^5\frac{4x^5+2x^3-5}{x^5}}=\lim_{x\rightarrow\infty}\frac{3x-\frac{1}{x}+\frac{2}{x^3}-\frac{3}{x^5}}{4+\frac{2}{x^2}-\frac{5}{x^5}}$$
The limit of the quotient is the quotient of limits:
$$\lim_{x\rightarrow\infty}\frac{3x-\frac{1}{x}+\frac{2}{x^3}-\frac{3}{x^5}}{4+\frac{2}{x^2}-\frac{5}{x^5}}=\frac{\lim_{x\rightarrow\infty}(3x-\frac{1}{x}+\frac{2}{x^3}-\frac{3}{x^5})}{\lim_{x\rightarrow\infty}(4+\frac{2}{x^2}-\frac{5}{x^5})}$$
The limit of a sum/difference is the sum/difference of limits: $$\frac{\lim_{x\rightarrow\infty}(3x-\frac{1}{x}+\frac{2}{x^3}-\frac{3}{x^5})}{\lim_{x\rightarrow\infty}(4+\frac{2}{x^2}-\frac{5}{x^5})}=\frac{(-\lim_{x\rightarrow\infty}\frac{3}{x^5}+\lim_{x\rightarrow\infty}\frac{2}{x^3}-\lim_{x\rightarrow\infty}\frac{1}{x}+\lim_{x\rightarrow\infty}3x)}{\lim_{x\rightarrow\infty}(4+\frac{2}{x^2}-\frac{5}{x^5})}$$
The limit of a quotient is the quotient of limits:
$$\frac{(-\lim_{x\rightarrow\infty}\frac{3}{x^5}+\lim_{x\rightarrow\infty}\frac{2}{x^3}-\lim_{x\rightarrow\infty}\frac{1}{x}+\lim_{x\rightarrow\infty}3x)}{\lim_{x\rightarrow\infty}(4+\frac{2}{x^2}-\frac{5}{x^5})}=\frac{(-\lim_{x\rightarrow\infty}\frac{3}{x^5}+\lim_{x\rightarrow\infty}\frac{2}{x^3}-\lim_{x\rightarrow\infty}3x)-\frac{\lim_{x\rightarrow\infty}1}{\lim_{x\rightarrow\infty}x}}{\lim_{x\rightarrow\infty}(4+\frac{2}{x^2}-\frac{5}{x^5})}$$
The limit of a constant is equal to the constant:
$$\frac{-\lim_{x\rightarrow\infty}\frac{3}{x^5}+\lim_{x\rightarrow\infty}\frac{2}{x^3}-\lim_{x\rightarrow\infty}3x-\frac{\lim_{x\rightarrow\infty}1}{\lim_{x\rightarrow\infty}x}}{\lim_{x\rightarrow\infty}(4+\frac{2}{x^2}-\frac{5}{x^5})}=\frac{-\lim_{x\rightarrow\infty}\frac{3}{x^5}+\lim_{x\rightarrow\infty}\frac{2}{x^3}-\lim_{x\rightarrow\infty}3x-\frac{1}{\lim_{x\rightarrow\infty}x}}{\lim_{x\rightarrow\infty}(4+\frac{2}{x^2}-\frac{5}{x^5})}$$
Constant divided by a very big number equals 0:
$$\frac{-\lim_{x\rightarrow\infty}\frac{3}{x^5}+\lim_{x\rightarrow\infty}\frac{2}{x^3}-\lim_{x\rightarrow\infty}3x-\frac{1}{\lim_{x\rightarrow\infty}x}}{\lim_{x\rightarrow\infty}(4+\frac{2}{x^2}-\frac{5}{x^5})}=\frac{-\lim_{x\rightarrow\infty}\frac{3}{x^5}+\lim_{x\rightarrow\infty}\frac{2}{x^3}-\lim_{x\rightarrow\infty}3x-(0)}{\lim_{x\rightarrow\infty}(4+\frac{2}{x^2}-\frac{5}{x^5})}$$
Apply the constant multiple rule $$\lim_{x\rightarrow\infty}cf(x)=c\cdot\lim_{x\rightarrow\infty}f(x)\ with\ c=3\ and\ f(x)=\frac{1}{x^5}$$
$$\frac{-\lim_{x\rightarrow\infty}\frac{3}{x^5}+\lim_{x\rightarrow\infty}\frac{2}{x^3}-\lim_{x\rightarrow\infty}3x}{\lim_{x\rightarrow\infty}(4+\frac{2}{x^2}-\frac{5}{x^5})}=\frac{\lim_{x\rightarrow\infty}\frac{2}{x^3}-\lim_{x\rightarrow\infty}3x-(3\lim_{x\rightarrow\infty}\frac{1}{x^5})}{\lim_{x\rightarrow\infty}(4+\frac{2}{x^2}-\frac{5}{x^5})}$$
The limit of a quotient is the quotient of limits:
$$\frac{\lim_{x\rightarrow\infty}\frac{2}{x^3}-\lim_{x\rightarrow\infty}3x-3\lim_{x\rightarrow\infty}\frac{1}{x^5}}{\lim_{x\rightarrow\infty}(4+\frac{2}{x^2}-\frac{5}{x^5})}=\frac{\lim_{x\rightarrow\infty}\frac{2}{x^3}-\lim_{x\rightarrow\infty}3x-3\frac{\lim_{x\rightarrow\infty}1}{\lim_{x\rightarrow\infty}x^5}}{\lim_{x\rightarrow\infty}(4+\frac{2}{x^2}-\frac{5}{x^5})}$$
The limit of a constant is equal to the constant:
$$\frac{\lim_{x\rightarrow\infty}\frac{2}{x^3}-\lim_{x\rightarrow\infty}3x-3\frac{\lim_{x\rightarrow\infty}1}{\lim_{x\rightarrow\infty}x^5}}{\lim_{x\rightarrow\infty}(4+\frac{2}{x^2}-\frac{5}{x^5})}=\frac{\lim_{x\rightarrow\infty}\frac{2}{x^3}-\lim_{x\rightarrow\infty}3x-\frac{3}{\lim_{x\rightarrow\infty}x^5}}{\lim_{x\rightarrow\infty}(4+\frac{2}{x^2}-\frac{5}{x^5})}$$
Constant divided by a very big number equals 0:
$$\frac{\lim_{x\rightarrow\infty}\frac{2}{x^3}-\lim_{x\rightarrow\infty}3x-\frac{3}{\lim_{x\rightarrow\infty}x^5}}{\lim_{x\rightarrow\infty}(4+\frac{2}{x^2}-\frac{5}{x^5})}=\frac{\lim_{x\rightarrow\infty}\frac{2}{x^3}-\lim_{x\rightarrow\infty}3x-3(0)}{\lim_{x\rightarrow\infty}(4+\frac{2}{x^2}-\frac{5}{x^5})}$$
Apply the constant multiple rule:
$$\frac{\lim_{x\rightarrow\infty}\frac{2}{x^3}-\lim_{x\rightarrow\infty}3x}{\lim_{x\rightarrow\infty}(4+\frac{2}{x^2}-\frac{5}{x^5})}=\frac{\lim_{x\rightarrow\infty}3x+(2\lim_{x\rightarrow\infty}\frac{1}{x^3})}{\lim_{x\rightarrow\infty}(4+\frac{2}{x^2}-\frac{5}{x^5})}$$
The limit of a quotient is the quotient of limits:
$$\frac{\lim_{x\rightarrow\infty}3x+(2\lim_{x\rightarrow\infty}\frac{1}{x^3})}{\lim_{x\rightarrow\infty}(4+\frac{2}{x^2}-\frac{5}{x^5})}=\frac{\lim_{x\rightarrow\infty}3x+2\frac{\lim_{x\rightarrow\infty}1}{\lim_{x\rightarrow}x^3}}{\lim_{x\rightarrow\infty}(4+\frac{2}{x^2}-\frac{5}{x^5})}$$
The limit of constant is equal to the constant:
$$\frac{\lim_{x\rightarrow\infty}3x+2\frac{\lim_{x\rightarrow\infty}1}{\lim_{x\rightarrow}x^3}}{\lim_{x\rightarrow\infty}(4+\frac{2}{x^2}-\frac{5}{x^5})}=\frac{\lim_{x\rightarrow\infty}3x+\frac{2}{\lim_{x\rightarrow}x^3}}{\lim_{x\rightarrow\infty}(4+\frac{2}{x^2}-\frac{5}{x^5})}$$
Constant divided by a very big number equals 0:
$$\frac{\lim_{x\rightarrow\infty}3x+\frac{2}{\lim_{x\rightarrow}x^3}}{\lim_{x\rightarrow\infty}(4+\frac{2}{x^2}-\frac{5}{x^5})}=\frac{\lim_{x\rightarrow\infty}3x+0}{\lim_{x\rightarrow\infty}(4+\frac{2}{x^2}-\frac{5}{x^5})}$$
Apply the constant multiple rule:
$$\frac{\lim_{x\rightarrow\infty}3x}{\lim_{x\rightarrow\infty}(4+\frac{2}{x^2}-\frac{5}{x^5})}=\frac{3\lim_{x\rightarrow\infty}x}{\lim_{x\rightarrow\infty}(4+\frac{2}{x^2}-\frac{5}{x^5})}$$
The function grows without a bound:
$$\lim_{x\rightarrow\infty}x=\infty$$
The limit of a sum/difference is the sum/difference of limits:
$$\infty\lim_{x\rightarrow\infty}(4+\frac{2}{x^2}-\frac{5}{x^5})^{-1}$$
$$=\infty(\lim_{x\rightarrow\infty}4+\lim_{x\rightarrow\infty}\frac{2}{x^2}-\lim_{x\rightarrow\infty}\frac{5}{x^5})^{-1}$$
Apply the constant multiple rule
$$\infty(4+\lim_{x\rightarrow\infty}\frac{2}{x^2}-\lim_{x\rightarrow\infty}\frac{5}{x^5})^{-1}$$
$$\infty(4+\lim_{x\rightarrow\infty}\frac{2}{x^2}-5\lim_{x\rightarrow\infty}\frac{1}{x^5})^{-1}$$
The limit of a qutient is the quotient of limits:
$$=\infty(4+\lim_{x\rightarrow\infty}\frac{2}{x^2}-5\frac{\lim_{x\rightarrow\infty}1}{\lim_{x\rightarrow\infty}x^5})^{-1}$$
The limit of a constant is equal to the constant:
$$=\infty(4+\lim_{x\rightarrow\infty}\frac{2}{x^2}-\frac{5}{\lim_{x\rightarrow\infty}x^5})^{-1}$$
Constant divided by a very big number equals 0:
$$=\infty(4+\lim_{x\rightarrow\infty}\frac{2}{x^2}-5(0))^{-1}$$
Apply the constant multiple rule:
$$=\infty(4+2\lim_{x\rightarrow\infty}\frac{1}{x^2})^{-1}$$
The limit of a constant is equal to the constant:
$$=\infty(4+2(0))^{-1}$$
Therefore,
$$\lim_{x\rightarrow\infty}\frac{3x^6-x^4+2x^2-3}{4x^5+2x^3-5}=\infty$$

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