What is the limit as x approaches infinity of (4^{x+1}+2^x)/(4^x) ?

The limit as x approaches infinity of (4^{x+1}+2^x)/(4^x) is 4

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limit as x approaches infinity of (4^{x+1}+2^x)/(4^x)

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

4

Solution steps

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

Factor out

4^{x + 1}

from

4^{x + 1} + 2^{x} : 4^{x + 1} (1 + \frac{2 ^{x}}{4 ^{x + 1}})

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

Simplify

\frac{4 ^{x + 1} ( 1 + \frac{2 ^{x}}{4 ^{x + 1}} )}{4 ^{x}} : 4 (\frac{2 ^{x}}{4 ^{x + 1}} + 1)

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

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

= 4 \cdot x \to \infty lim (\frac{2 ^{x}}{4 ^{x + 1}} + 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

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

x \to \infty lim (\frac{2 ^{x}}{4 ^{x + 1}}) = 0

x \to \infty lim (1) = 1

= 4 (0 + 1)

Simplify

= 4

\frac{d}{d x} (- 7 ln (5 x^{2} - 11))

\int_{2}^{3} \frac{1}{- 7 + 16 x - 4 x ^{2}} d x

\int 1 0^{x^{2}} x d x

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

x \to - 2 - lim (f (x) x)

What is the limit as x approaches infinity of (4^{x+1}+2^x)/(4^x) ?
The limit as x approaches infinity of (4^{x+1}+2^x)/(4^x) is 4