# Limit Calculator

## Calculate limits step by step

This free calculator will try to find the limit (two-sided or one-sided, including left and right) of the given function at the given point (including infinity), with steps shown.

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### Solution

Your input: find $\lim_{x \to -\infty} \frac{2 x^{3} + 15 x^{2} + 22 x - 11}{x^{2} + 8 x + 15}$

Multiply and divide by $x^{2}$:

$$\color{red}{\lim_{x \to -\infty} \frac{2 x^{3} + 15 x^{2} + 22 x - 11}{x^{2} + 8 x + 15}} = \color{red}{\lim_{x \to -\infty} \frac{x^{2} \frac{2 x^{3} + 15 x^{2} + 22 x - 11}{x^{2}}}{x^{2} \frac{x^{2} + 8 x + 15}{x^{2}}}}$$

Divide:

$$\color{red}{\lim_{x \to -\infty} \frac{x^{2} \frac{2 x^{3} + 15 x^{2} + 22 x - 11}{x^{2}}}{x^{2} \frac{x^{2} + 8 x + 15}{x^{2}}}} = \color{red}{\lim_{x \to -\infty} \frac{2 x + 15 + \frac{22}{x} - \frac{11}{x^{2}}}{1 + \frac{8}{x} + \frac{15}{x^{2}}}}$$

The limit of the quotient is the quotient of limits:

$$\color{red}{\lim_{x \to -\infty} \frac{2 x + 15 + \frac{22}{x} - \frac{11}{x^{2}}}{1 + \frac{8}{x} + \frac{15}{x^{2}}}} = \color{red}{\frac{\lim_{x \to -\infty}\left(2 x + 15 + \frac{22}{x} - \frac{11}{x^{2}}\right)}{\lim_{x \to -\infty}\left(1 + \frac{8}{x} + \frac{15}{x^{2}}\right)}}$$

The limit of a sum/difference is the sum/difference of limits:

$$\frac{\color{red}{\lim_{x \to -\infty}\left(2 x + 15 + \frac{22}{x} - \frac{11}{x^{2}}\right)}}{\lim_{x \to -\infty}\left(1 + \frac{8}{x} + \frac{15}{x^{2}}\right)} = \frac{\color{red}{\left(\lim_{x \to -\infty} 15 - \lim_{x \to -\infty} \frac{11}{x^{2}} + \lim_{x \to -\infty} \frac{22}{x} + \lim_{x \to -\infty} 2 x\right)}}{\lim_{x \to -\infty}\left(1 + \frac{8}{x} + \frac{15}{x^{2}}\right)}$$

The limit of a constant is equal to the constant:

$$\frac{- \lim_{x \to -\infty} \frac{11}{x^{2}} + \lim_{x \to -\infty} \frac{22}{x} + \lim_{x \to -\infty} 2 x + \color{red}{\lim_{x \to -\infty} 15}}{\lim_{x \to -\infty}\left(1 + \frac{8}{x} + \frac{15}{x^{2}}\right)} = \frac{- \lim_{x \to -\infty} \frac{11}{x^{2}} + \lim_{x \to -\infty} \frac{22}{x} + \lim_{x \to -\infty} 2 x + \color{red}{\left(15\right)}}{\lim_{x \to -\infty}\left(1 + \frac{8}{x} + \frac{15}{x^{2}}\right)}$$

Apply the constant multiple rule $\lim_{x \to -\infty} c f{\left(x \right)} = c \lim_{x \to -\infty} f{\left(x \right)}$ with $c=11$ and $f{\left(x \right)} = \frac{1}{x^{2}}$:

$$\frac{\lim_{x \to -\infty} \frac{22}{x} + \lim_{x \to -\infty} 2 x + 15 - \color{red}{\lim_{x \to -\infty} \frac{11}{x^{2}}}}{\lim_{x \to -\infty}\left(1 + \frac{8}{x} + \frac{15}{x^{2}}\right)} = \frac{\lim_{x \to -\infty} \frac{22}{x} + \lim_{x \to -\infty} 2 x + 15 - \color{red}{\left(11 \lim_{x \to -\infty} \frac{1}{x^{2}}\right)}}{\lim_{x \to -\infty}\left(1 + \frac{8}{x} + \frac{15}{x^{2}}\right)}$$

The limit of a quotient is the quotient of limits:

$$\frac{\lim_{x \to -\infty} \frac{22}{x} + \lim_{x \to -\infty} 2 x + 15 - 11 \color{red}{\lim_{x \to -\infty} \frac{1}{x^{2}}}}{\lim_{x \to -\infty}\left(1 + \frac{8}{x} + \frac{15}{x^{2}}\right)} = \frac{\lim_{x \to -\infty} \frac{22}{x} + \lim_{x \to -\infty} 2 x + 15 - 11 \color{red}{\frac{\lim_{x \to -\infty} 1}{\lim_{x \to -\infty} x^{2}}}}{\lim_{x \to -\infty}\left(1 + \frac{8}{x} + \frac{15}{x^{2}}\right)}$$

The limit of a constant is equal to the constant:

$$\frac{\lim_{x \to -\infty} \frac{22}{x} + \lim_{x \to -\infty} 2 x + 15 - \frac{11 \color{red}{\lim_{x \to -\infty} 1}}{\lim_{x \to -\infty} x^{2}}}{\lim_{x \to -\infty}\left(1 + \frac{8}{x} + \frac{15}{x^{2}}\right)} = \frac{\lim_{x \to -\infty} \frac{22}{x} + \lim_{x \to -\infty} 2 x + 15 - \frac{11 \color{red}{1}}{\lim_{x \to -\infty} x^{2}}}{\lim_{x \to -\infty}\left(1 + \frac{8}{x} + \frac{15}{x^{2}}\right)}$$

Constant divided by a very big number equals $0$:

$$\frac{\lim_{x \to -\infty} \frac{22}{x} + \lim_{x \to -\infty} 2 x + 15 - 11 \color{red}{1 \frac{1}{\lim_{x \to -\infty} x^{2}}}}{\lim_{x \to -\infty}\left(1 + \frac{8}{x} + \frac{15}{x^{2}}\right)} = \frac{\lim_{x \to -\infty} \frac{22}{x} + \lim_{x \to -\infty} 2 x + 15 - 11 \color{red}{\left(0\right)}}{\lim_{x \to -\infty}\left(1 + \frac{8}{x} + \frac{15}{x^{2}}\right)}$$

Apply the constant multiple rule $\lim_{x \to -\infty} c f{\left(x \right)} = c \lim_{x \to -\infty} f{\left(x \right)}$ with $c=22$ and $f{\left(x \right)} = \frac{1}{x}$:

$$\frac{\lim_{x \to -\infty} 2 x + 15 + \color{red}{\lim_{x \to -\infty} \frac{22}{x}}}{\lim_{x \to -\infty}\left(1 + \frac{8}{x} + \frac{15}{x^{2}}\right)} = \frac{\lim_{x \to -\infty} 2 x + 15 + \color{red}{\left(22 \lim_{x \to -\infty} \frac{1}{x}\right)}}{\lim_{x \to -\infty}\left(1 + \frac{8}{x} + \frac{15}{x^{2}}\right)}$$

The limit of a quotient is the quotient of limits:

$$\frac{\lim_{x \to -\infty} 2 x + 15 + 22 \color{red}{\lim_{x \to -\infty} \frac{1}{x}}}{\lim_{x \to -\infty}\left(1 + \frac{8}{x} + \frac{15}{x^{2}}\right)} = \frac{\lim_{x \to -\infty} 2 x + 15 + 22 \color{red}{\frac{\lim_{x \to -\infty} 1}{\lim_{x \to -\infty} x}}}{\lim_{x \to -\infty}\left(1 + \frac{8}{x} + \frac{15}{x^{2}}\right)}$$

The limit of a constant is equal to the constant:

$$\frac{\lim_{x \to -\infty} 2 x + 15 + \frac{22 \color{red}{\lim_{x \to -\infty} 1}}{\lim_{x \to -\infty} x}}{\lim_{x \to -\infty}\left(1 + \frac{8}{x} + \frac{15}{x^{2}}\right)} = \frac{\lim_{x \to -\infty} 2 x + 15 + \frac{22 \color{red}{1}}{\lim_{x \to -\infty} x}}{\lim_{x \to -\infty}\left(1 + \frac{8}{x} + \frac{15}{x^{2}}\right)}$$

Constant divided by a very big number equals $0$:

$$\frac{\lim_{x \to -\infty} 2 x + 15 + 22 \color{red}{1 \frac{1}{\lim_{x \to -\infty} x}}}{\lim_{x \to -\infty}\left(1 + \frac{8}{x} + \frac{15}{x^{2}}\right)} = \frac{\lim_{x \to -\infty} 2 x + 15 + 22 \color{red}{\left(0\right)}}{\lim_{x \to -\infty}\left(1 + \frac{8}{x} + \frac{15}{x^{2}}\right)}$$

Apply the constant multiple rule $\lim_{x \to -\infty} c f{\left(x \right)} = c \lim_{x \to -\infty} f{\left(x \right)}$ with $c=2$ and $f{\left(x \right)} = x$:

$$\frac{15 + \color{red}{\lim_{x \to -\infty} 2 x}}{\lim_{x \to -\infty}\left(1 + \frac{8}{x} + \frac{15}{x^{2}}\right)} = \frac{15 + \color{red}{\left(2 \lim_{x \to -\infty} x\right)}}{\lim_{x \to -\infty}\left(1 + \frac{8}{x} + \frac{15}{x^{2}}\right)}$$

The function decreases without a bound:

$$\lim_{x \to -\infty} x = -\infty$$

The limit of a sum/difference is the sum/difference of limits:

$$- \infty \color{red}{\lim_{x \to -\infty}\left(1 + \frac{8}{x} + \frac{15}{x^{2}}\right)}^{-1} = - \infty \color{red}{\left(\lim_{x \to -\infty} 1 + \lim_{x \to -\infty} \frac{15}{x^{2}} + \lim_{x \to -\infty} \frac{8}{x}\right)}^{-1}$$

The limit of a constant is equal to the constant:

$$- \infty \left(\lim_{x \to -\infty} \frac{15}{x^{2}} + \lim_{x \to -\infty} \frac{8}{x} + \color{red}{\lim_{x \to -\infty} 1}\right)^{-1} = - \infty \left(\lim_{x \to -\infty} \frac{15}{x^{2}} + \lim_{x \to -\infty} \frac{8}{x} + \color{red}{1}\right)^{-1}$$

Apply the constant multiple rule $\lim_{x \to -\infty} c f{\left(x \right)} = c \lim_{x \to -\infty} f{\left(x \right)}$ with $c=8$ and $f{\left(x \right)} = \frac{1}{x}$:

$$- \infty \left(\lim_{x \to -\infty} \frac{15}{x^{2}} + 1 + \color{red}{\lim_{x \to -\infty} \frac{8}{x}}\right)^{-1} = - \infty \left(\lim_{x \to -\infty} \frac{15}{x^{2}} + 1 + \color{red}{\left(8 \lim_{x \to -\infty} \frac{1}{x}\right)}\right)^{-1}$$

The limit of a quotient is the quotient of limits:

$$- \infty \left(\lim_{x \to -\infty} \frac{15}{x^{2}} + 1 + 8 \color{red}{\lim_{x \to -\infty} \frac{1}{x}}\right)^{-1} = - \infty \left(\lim_{x \to -\infty} \frac{15}{x^{2}} + 1 + 8 \color{red}{\frac{\lim_{x \to -\infty} 1}{\lim_{x \to -\infty} x}}\right)^{-1}$$

The limit of a constant is equal to the constant:

$$- \infty \left(\lim_{x \to -\infty} \frac{15}{x^{2}} + 1 + \frac{8 \color{red}{\lim_{x \to -\infty} 1}}{\lim_{x \to -\infty} x}\right)^{-1} = - \infty \left(\lim_{x \to -\infty} \frac{15}{x^{2}} + 1 + \frac{8 \color{red}{1}}{\lim_{x \to -\infty} x}\right)^{-1}$$

Constant divided by a very big number equals $0$:

$$- \infty \left(\lim_{x \to -\infty} \frac{15}{x^{2}} + 1 + 8 \color{red}{1 \frac{1}{\lim_{x \to -\infty} x}}\right)^{-1} = - \infty \left(\lim_{x \to -\infty} \frac{15}{x^{2}} + 1 + 8 \color{red}{\left(0\right)}\right)^{-1}$$

Apply the constant multiple rule $\lim_{x \to -\infty} c f{\left(x \right)} = c \lim_{x \to -\infty} f{\left(x \right)}$ with $c=15$ and $f{\left(x \right)} = \frac{1}{x^{2}}$:

$$- \infty \left(1 + \color{red}{\lim_{x \to -\infty} \frac{15}{x^{2}}}\right)^{-1} = - \infty \left(1 + \color{red}{\left(15 \lim_{x \to -\infty} \frac{1}{x^{2}}\right)}\right)^{-1}$$

The limit of a quotient is the quotient of limits:

$$- \infty \left(1 + 15 \color{red}{\lim_{x \to -\infty} \frac{1}{x^{2}}}\right)^{-1} = - \infty \left(1 + 15 \color{red}{\frac{\lim_{x \to -\infty} 1}{\lim_{x \to -\infty} x^{2}}}\right)^{-1}$$

The limit of a constant is equal to the constant:

$$- \infty \left(1 + \frac{15 \color{red}{\lim_{x \to -\infty} 1}}{\lim_{x \to -\infty} x^{2}}\right)^{-1} = - \infty \left(1 + \frac{15 \color{red}{1}}{\lim_{x \to -\infty} x^{2}}\right)^{-1}$$

Constant divided by a very big number equals $0$:

$$- \infty \left(1 + 15 \color{red}{1 \frac{1}{\lim_{x \to -\infty} x^{2}}}\right)^{-1} = - \infty \left(1 + 15 \color{red}{\left(0\right)}\right)^{-1}$$

Therefore,

$$\lim_{x \to -\infty} \frac{2 x^{3} + 15 x^{2} + 22 x - 11}{x^{2} + 8 x + 15} = -\infty$$

Answer: $\lim_{x \to -\infty} \frac{2 x^{3} + 15 x^{2} + 22 x - 11}{x^{2} + 8 x + 15}=-\infty$