What is the limit as x approaches infinity of-1+e^{(x+1)} ?

The limit as x approaches infinity of-1+e^{(x+1)} is infinity

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limit as x approaches infinity of-1+e^{(x+1)}

x \to \infty lim (- 1 + e^{(x + 1)})

\infty

Solution steps

x \to \infty lim (- 1 + e^{x + 1})

x \to a lim [f (x) \pm g (x)] = x \to a lim f (x) \pm x \to a lim g (x)

With the exception of indeterminate form

= - x \to \infty lim (1) + x \to \infty lim (e^{x + 1})

x \to \infty lim (1) = 1

x \to \infty lim (e^{x + 1}) = \infty

= - 1 + \infty

Apply Infinity Property:

- c + \infty = \infty

= \infty

x \to \infty lim ((6)^{x})

\int \frac{x ^{4} e ^{x} - 4 x ^{3}}{x ^{4}} d x

y^{^{'}} = 9.8 - 0.2 y, y (0) = 100

n = - 4 \sum \infty (- \frac{4}{9})^{n}

x \to 5 + lim (\frac{x - 5}{∣ x - 5 ∣})

What is the limit as x approaches infinity of-1+e^{(x+1)} ?
The limit as x approaches infinity of-1+e^{(x+1)} is infinity