What is the limit as x approaches-infinity of 2^x+1 ?

The limit as x approaches-infinity of 2^x+1 is 1

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limit as x approaches-infinity of 2^x+1

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

1

Solution steps

x \to - \infty lim (2^{x} + 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

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

x \to - \infty lim (2^{x}) = 0

x \to - \infty lim (1) = 1

= 0 + 1

Simplify

= 1

\int \frac{e ^{5 x} + 5 e ^{x} + 7}{e ^{x}} d x

f (x) = 5 cot (x), at x = \frac{π}{4}

\frac{d y}{d x}, x + y = x^{2}

y^{^{''}} - 6 y^{^{'}} - 27 y = 0

\int \frac{1}{y - x} d y

What is the limit as x approaches-infinity of 2^x+1 ?
The limit as x approaches-infinity of 2^x+1 is 1