Limit Calculator

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.

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$\displaystyle{\large{\infty}}$ , type inf.

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Solution

Your input: find $\lim_{x \to -\infty}\left(x + e^{- x}\right)$

Multiply and divide by $e^{- x}$ :

${\color{red}{\lim_{x \to -\infty}\left(x + e^{- x}\right)}} = {\color{red}{\lim_{x \to -\infty} \left(x + e^{- x}\right) e^{x} e^{- x}}}$

Divide:

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

The limit of a product/quotient is the product/quotient of limits:

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

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

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

The limit of a constant is equal to the constant:

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

Rewrite:

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

Since we have an indeterminate form of type $\frac{\infty}{\infty}$ , we can apply the l'Hopital's rule:

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

For steps, see derivative calculator.

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

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

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

Move the limit under the exponential:

$\lim_{x \to -\infty} e^{- x} \left(1 - {\color{red}{\lim_{x \to -\infty} e^{x}}}\right) = \lim_{x \to -\infty} e^{- x} \left(1 - {\color{red}{e^{\lim_{x \to -\infty} x}}}\right)$

The function decreases without a bound:

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

The function grows without a bound:

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

Therefore,

$\lim_{x \to -\infty}\left(x + e^{- x}\right) = \infty$

Answer: $\lim_{x \to -\infty}\left(x + e^{- x}\right)=\infty$

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Solution