Calculadora de limite
Calcular limites passo a passo
Esta calculadora gratuita tentará encontrar o limite (bilateral ou unilateral, incluindo esquerda e direita) da função dada no ponto dado (incluindo infinito), com passos mostrados.
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: $$$\lim_{x \to -\infty} \frac{2 x^{3} + 15 x^{2} + 22 x - 11}{x^{2} + 8 x + 15}=-\infty$$$