What is the limit as x approaches infinity of (10-6x^2)/(5+3e^x) ?

The limit as x approaches infinity of (10-6x^2)/(5+3e^x) is 0

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limit as x approaches infinity of (10-6x^2)/(5+3e^x)

x \to \infty lim (\frac{10 - 6 x ^{2}}{5 + 3 e ^{x}})

0

Solution steps

x \to \infty lim (\frac{10 - 6 x ^{2}}{5 + 3 e ^{x}})

Apply L'Hopital's Rule

: x \to \infty lim (\frac{- 12 x}{3 e ^{x}})

= x \to \infty lim (\frac{- 12 x}{3 e ^{x}})

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

= \frac{- 12}{3} \cdot x \to \infty lim (\frac{x}{e ^{x}})

Simplify

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

Apply L'Hopital's Rule

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

= (- 4) \cdot x \to \infty lim (\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

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

x \to \infty lim (1) = 1

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

= (- 4) \frac{1}{\infty}

Simplify

(- 4) \frac{1}{\infty} : 0

= 0

x \to - 2 - lim (\frac{x - 2}{x ^{2} - 4})

x \to - \frac{a}{2} lim (4 x^{2} - 3 x + 1)

x \to 0 + lim (- 1 + ln (x) x)

x \to - 3 + lim (\frac{x + 1}{x ^{2} - 9})

x \to 4 lim (\frac{x - 4}{4 - x})

What is the limit as x approaches infinity of (10-6x^2)/(5+3e^x) ?
The limit as x approaches infinity of (10-6x^2)/(5+3e^x) is 0