What is the limit as x approaches infinity of (x^3+2x+4)/(e^{1x)+1} ?

The limit as x approaches infinity of (x^3+2x+4)/(e^{1x)+1} is 0

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limit as x approaches infinity of (x^3+2x+4)/(e^{1x)+1}

x \to \infty lim (\frac{x ^{3} + 2 x + 4}{e ^{1 x} + 1})

0

Solution steps

x \to \infty lim (\frac{x ^{3} + 2 x + 4}{e ^{1 \cdot x} + 1})

Divide by

e^{1 x} : \frac{\frac{x ^{3}}{e ^{x}} + \frac{2 x}{e ^{x}} + \frac{4}{e ^{x}}}{1 + \frac{1}{e ^{x}}}

= x \to \infty lim (\frac{\frac{x ^{3}}{e ^{x}} + \frac{2 x}{e ^{x}} + \frac{4}{e ^{x}}}{1 + \frac{1}{e ^{x}}})

x \to a lim [\frac{f ( x )}{g ( x )}] = \frac{lim _{x \to a} f ( x )}{lim _{x \to a} g ( x )}, x \to a lim g (x) \neq = 0

With the exception of indeterminate form

= \frac{lim _{x \to \infty} ( \frac{x ^{3}}{e ^{x}} + \frac{2 x}{e ^{x}} + \frac{4}{e ^{x}} )}{lim _{x \to \infty} ( 1 + \frac{1}{e ^{x}} )}

x \to \infty lim (\frac{x ^{3}}{e ^{x}} + \frac{2 x}{e ^{x}} + \frac{4}{e ^{x}}) = 0

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

= \frac{0}{1}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

y = 4 x^{2} - 6 x, at x = 2

\frac{d y}{d x} = \frac{y ^{2} + x ^{2}}{y x}

\int cos^{9} (x) d x

\int (8 x^{3} - 3 x^{2} + 8) d x

\frac{d}{d x} (x^{5} e^{6 x})

What is the limit as x approaches infinity of (x^3+2x+4)/(e^{1x)+1} ?
The limit as x approaches infinity of (x^3+2x+4)/(e^{1x)+1} is 0