Limit Calculator

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