What is the limit as x approaches infinity of ((e^{2x}-8-19x))/((19x)^2) ?

The limit as x approaches infinity of ((e^{2x}-8-19x))/((19x)^2) is infinity

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limit as x approaches infinity of ((e^{2x}-8-19x))/((19x)^2)

x \to \infty lim (\frac{( e ^{2 x} - 8 - 19 x )}{( 19 x ) ^{2}})

\infty

Solution steps

x \to \infty lim (\frac{e ^{2 x} - 8 - 19 x}{( 19 x ) ^{2}})

Expand

\frac{e ^{2 x} - 8 - 19 x}{( 19 x ) ^{2}} : \frac{e ^{2 x}}{361 x ^{2}} - \frac{8}{361 x ^{2}} - \frac{1}{19 x}

= x \to \infty lim (\frac{e ^{2 x}}{361 x ^{2}} - \frac{8}{361 x ^{2}} - \frac{1}{19 x})

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 (\frac{e ^{2 x}}{361 x ^{2}}) - x \to \infty lim (\frac{8}{361 x ^{2}}) - x \to \infty lim (\frac{1}{19 x})

x \to \infty lim (\frac{e ^{2 x}}{361 x ^{2}}) = \infty

x \to \infty lim (\frac{8}{361 x ^{2}}) = 0

x \to \infty lim (\frac{1}{19 x}) = 0

= \infty - 0 - 0

Simplify

\infty - 0 - 0 : \infty

= \infty

y^{^{''}} + 49 y = 0

\int 64 sin^{2} (x) d x

f (x) = x^{3} - 11 x, at x = 3

n = 2 \sum \infty (\frac{3}{4})^{n}

\frac{d}{d x} ((4 x^{5} - 2 x) (3 x + e^{x}))

What is the limit as x approaches infinity of ((e^{2x}-8-19x))/((19x)^2) ?
The limit as x approaches infinity of ((e^{2x}-8-19x))/((19x)^2) is infinity