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.
Different techniques are used to handle limits (including indeterminate forms): limit laws, rewriting and simplifying, L'Hôpital's rule, rationalizing the denominator, taking natural logarithm, etc.
Solution
Your input: find $$$\lim_{x \to \infty} \frac{x^{2} + 2}{x^{2} - 4}$$$
Multiply and divide by $$$x^{2}$$$:
$${\color{red}{\lim_{x \to \infty} \frac{x^{2} + 2}{x^{2} - 4}}} = {\color{red}{\lim_{x \to \infty} \frac{x^{2} \frac{x^{2} + 2}{x^{2}}}{x^{2} \frac{x^{2} - 4}{x^{2}}}}}$$
Divide:
$${\color{red}{\lim_{x \to \infty} \frac{x^{2} \frac{x^{2} + 2}{x^{2}}}{x^{2} \frac{x^{2} - 4}{x^{2}}}}} = {\color{red}{\lim_{x \to \infty} \frac{1 + \frac{2}{x^{2}}}{1 - \frac{4}{x^{2}}}}}$$
The limit of the quotient is the quotient of limits:
$${\color{red}{\lim_{x \to \infty} \frac{1 + \frac{2}{x^{2}}}{1 - \frac{4}{x^{2}}}}} = {\color{red}{\frac{\lim_{x \to \infty}\left(1 + \frac{2}{x^{2}}\right)}{\lim_{x \to \infty}\left(1 - \frac{4}{x^{2}}\right)}}}$$
The limit of a sum/difference is the sum/difference of limits:
$$\frac{{\color{red}{\lim_{x \to \infty}\left(1 + \frac{2}{x^{2}}\right)}}}{\lim_{x \to \infty}\left(1 - \frac{4}{x^{2}}\right)} = \frac{{\color{red}{\left(\lim_{x \to \infty} 1 + \lim_{x \to \infty} \frac{2}{x^{2}}\right)}}}{\lim_{x \to \infty}\left(1 - \frac{4}{x^{2}}\right)}$$
The limit of a constant is equal to the constant:
$$\frac{\lim_{x \to \infty} \frac{2}{x^{2}} + {\color{red}{\lim_{x \to \infty} 1}}}{\lim_{x \to \infty}\left(1 - \frac{4}{x^{2}}\right)} = \frac{\lim_{x \to \infty} \frac{2}{x^{2}} + {\color{red}{1}}}{\lim_{x \to \infty}\left(1 - \frac{4}{x^{2}}\right)}$$
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 $$$f{\left(x \right)} = \frac{1}{x^{2}}$$$:
$$\frac{1 + {\color{red}{\lim_{x \to \infty} \frac{2}{x^{2}}}}}{\lim_{x \to \infty}\left(1 - \frac{4}{x^{2}}\right)} = \frac{1 + {\color{red}{\left(2 \lim_{x \to \infty} \frac{1}{x^{2}}\right)}}}{\lim_{x \to \infty}\left(1 - \frac{4}{x^{2}}\right)}$$
The limit of a quotient is the quotient of limits:
$$\frac{1 + 2 {\color{red}{\lim_{x \to \infty} \frac{1}{x^{2}}}}}{\lim_{x \to \infty}\left(1 - \frac{4}{x^{2}}\right)} = \frac{1 + 2 {\color{red}{\frac{\lim_{x \to \infty} 1}{\lim_{x \to \infty} x^{2}}}}}{\lim_{x \to \infty}\left(1 - \frac{4}{x^{2}}\right)}$$
The limit of a constant is equal to the constant:
$$\frac{1 + \frac{2 {\color{red}{\lim_{x \to \infty} 1}}}{\lim_{x \to \infty} x^{2}}}{\lim_{x \to \infty}\left(1 - \frac{4}{x^{2}}\right)} = \frac{1 + \frac{2 {\color{red}{1}}}{\lim_{x \to \infty} x^{2}}}{\lim_{x \to \infty}\left(1 - \frac{4}{x^{2}}\right)}$$
Constant divided by a very big number equals $$$0$$$:
$$\frac{1 + 2 {\color{red}{1 \frac{1}{\lim_{x \to \infty} x^{2}}}}}{\lim_{x \to \infty}\left(1 - \frac{4}{x^{2}}\right)} = \frac{1 + 2 {\color{red}{\left(0\right)}}}{\lim_{x \to \infty}\left(1 - \frac{4}{x^{2}}\right)}$$
The limit of a sum/difference is the sum/difference of limits:
$${\color{red}{\lim_{x \to \infty}\left(1 - \frac{4}{x^{2}}\right)}}^{-1} = {\color{red}{\left(\lim_{x \to \infty} 1 - \lim_{x \to \infty} \frac{4}{x^{2}}\right)}}^{-1}$$
The limit of a constant is equal to the constant:
$$\left(- \lim_{x \to \infty} \frac{4}{x^{2}} + {\color{red}{\lim_{x \to \infty} 1}}\right)^{-1} = \left(- \lim_{x \to \infty} \frac{4}{x^{2}} + {\color{red}{1}}\right)^{-1}$$
Apply the constant multiple rule $$$\lim_{x \to \infty} c f{\left(x \right)} = c \lim_{x \to \infty} f{\left(x \right)}$$$ with $$$c=4$$$ and $$$f{\left(x \right)} = \frac{1}{x^{2}}$$$:
$$\left(1 - {\color{red}{\lim_{x \to \infty} \frac{4}{x^{2}}}}\right)^{-1} = \left(1 - {\color{red}{\left(4 \lim_{x \to \infty} \frac{1}{x^{2}}\right)}}\right)^{-1}$$
The limit of a quotient is the quotient of limits:
$$\left(1 - 4 {\color{red}{\lim_{x \to \infty} \frac{1}{x^{2}}}}\right)^{-1} = \left(1 - 4 {\color{red}{\frac{\lim_{x \to \infty} 1}{\lim_{x \to \infty} x^{2}}}}\right)^{-1}$$
The limit of a constant is equal to the constant:
$$\left(1 - \frac{4 {\color{red}{\lim_{x \to \infty} 1}}}{\lim_{x \to \infty} x^{2}}\right)^{-1} = \left(1 - \frac{4 {\color{red}{1}}}{\lim_{x \to \infty} x^{2}}\right)^{-1}$$
Constant divided by a very big number equals $$$0$$$:
$$\left(1 - 4 {\color{red}{1 \frac{1}{\lim_{x \to \infty} x^{2}}}}\right)^{-1} = \left(1 - 4 {\color{red}{\left(0\right)}}\right)^{-1}$$
Therefore,
$$\lim_{x \to \infty} \frac{x^{2} + 2}{x^{2} - 4} = 1$$
Answer: $$$\lim_{x \to \infty} \frac{x^{2} + 2}{x^{2} - 4}=1$$$