What is the limit as x approaches infinity of (e^{5x})/(9x^4+31x+1) ?

The limit as x approaches infinity of (e^{5x})/(9x^4+31x+1) is infinity

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limit as x approaches infinity of (e^{5x})/(9x^4+31x+1)

x \to \infty lim (\frac{e ^{5 x}}{9 x ^{4} + 31 x + 1})

\infty

Solution steps

x \to \infty lim (\frac{e ^{5 x}}{9 x ^{4} + 31 x + 1})

Divide by highest denominator power:

\frac{\frac{e ^{5 x}}{x ^{4}}}{9 + \frac{31}{x ^{3}} + \frac{1}{x ^{4}}}

= x \to \infty lim (\frac{\frac{e ^{5 x}}{x ^{4}}}{9 + \frac{31}{x ^{3}} + \frac{1}{x ^{4}}})

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{e ^{5 x}}{x ^{4}} )}{lim _{x \to \infty} ( 9 + \frac{31}{x ^{3}} + \frac{1}{x ^{4}} )}

x \to \infty lim (\frac{e ^{5 x}}{x ^{4}}) = \infty

x \to \infty lim (9 + \frac{31}{x ^{3}} + \frac{1}{x ^{4}}) = 9

= \frac{\infty}{9}

Apply Infinity Property:

\frac{\infty}{c} = \infty

= \infty

\frac{d}{d x} (sec^{2} (2 x + 4) \cdot 2)

f (x) = \frac{f ( x )}{x ^{7}}

5 x^{3} + 2 x^{2} + 3

x \to \infty lim (co x (x))

f (x) = x^{2} + x - 3, at x = 4

What is the limit as x approaches infinity of (e^{5x})/(9x^4+31x+1) ?
The limit as x approaches infinity of (e^{5x})/(9x^4+31x+1) is infinity