What is the limit as n approaches infinity of (-8+1/(2^n))/(7/(2^n)) ?

The limit as n approaches infinity of (-8+1/(2^n))/(7/(2^n)) is -infinity

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limit as n approaches infinity of (-8+1/(2^n))/(7/(2^n))

n \to \infty lim (\frac{- 8 + \frac{1}{2 ^{n}}}{\frac{7}{2 ^{n}}})

- \infty

Solution steps

n \to \infty lim (\frac{- 8 + \frac{1}{2 ^{n}}}{\frac{7}{2 ^{n}}})

Simplify

\frac{- 8 + \frac{1}{2 ^{n}}}{\frac{7}{2 ^{n}}} : \frac{1}{7} (- 2^{3 + n} + 1)

= n \to \infty lim (\frac{1}{7} (- 2^{3 + n} + 1))

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

= \frac{1}{7} \cdot n \to \infty lim (- 2^{3 + n} + 1)

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

= \frac{1}{7} (- n \to \infty lim (2^{3 + n}) + n \to \infty lim (1))

n \to \infty lim (2^{3 + n}) = \infty

n \to \infty lim (1) = 1

= \frac{1}{7} (- \infty + 1)

Simplify

\frac{1}{7} (- \infty + 1) : - \infty

= - \infty

x \to \infty lim (\frac{3 x ^{8} - 17 x ^{3} + 5}{8 x ^{4} + 12 x ^{3} - x})

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

x \to 0 lim (\frac{∣ x ∣}{3 x})

x \to 4 lim (\frac{\frac{2}{x + 1} - \frac{4}{x + 6}}{x + 4})

x \to 0 lim (\frac{2}{- sin ( x )})

What is the limit as n approaches infinity of (-8+1/(2^n))/(7/(2^n)) ?
The limit as n approaches infinity of (-8+1/(2^n))/(7/(2^n)) is -infinity