What is the limit as x approaches infinity of (10+9e^{(2x)})/(11-8e^{(3x))} ?

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limit as x approaches infinity of (10+9e^{(2x)})/(11-8e^{(3x))}

x \to \infty lim (\frac{10 + 9 e ^{(2 x)}}{11 - 8 e ^{(3 x)}})

0

Solution steps

x \to \infty lim (\frac{10 + 9 e ^{2 x}}{11 - 8 e ^{3 x}})

Divide by

e^{3 x} : \frac{\frac{10}{e ^{3 x}} + \frac{9}{e ^{x}}}{\frac{11}{e ^{3 x}} - 8}

= x \to \infty lim (\frac{\frac{10}{e ^{3 x}} + \frac{9}{e ^{x}}}{\frac{11}{e ^{3 x}} - 8})

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{10}{e ^{3 x}} + \frac{9}{e ^{x}} )}{lim _{x \to \infty} ( \frac{11}{e ^{3 x}} - 8 )}

x \to \infty lim (\frac{10}{e ^{3 x}} + \frac{9}{e ^{x}}) = 0

x \to \infty lim (\frac{11}{e ^{3 x}} - 8) = - 8

= \frac{0}{- 8}

Apply rule

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

= 0

\frac{d}{d x} (a r (x) cos (x))

\int_{0}^{5} x d x

ln (9 sin^{2} (x))

\frac{d}{d u} (\frac{5 u}{4 u ^{2} + 1})

y = \frac{1}{x}, at x = 2

What is the limit as x approaches infinity of (10+9e^{(2x)})/(11-8e^{(3x))} ?
The limit as x approaches infinity of (10+9e^{(2x)})/(11-8e^{(3x))} is 0