What is the limit as x approaches infinity of (cosh(2x))/(e^{2x)} ?

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

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limit as x approaches infinity of (cosh(2x))/(e^{2x)}

x \to \infty lim (\frac{cosh ( 2 x )}{e ^{2 x}})

\frac{1}{2}

Decimal

0.5

Solution steps

x \to \infty lim (\frac{cosh ( 2 x )}{e ^{2 x}})

Use the following identity:

cosh (x) = \frac{e ^{- x} + e ^{x}}{2}

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

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

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

Simplify

\frac{e ^{- 2 x} + e ^{2 x}}{e ^{2 x}} : \frac{1 + e ^{4 x}}{e ^{4 x}}

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

Divide by

e^{4 x} : \frac{1}{e ^{4 x}} + 1

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

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

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

x \to \infty lim (1) = 1

= \frac{1}{2} (0 + 1)

Simplify

= \frac{1}{2}

x \to 0 lim (\frac{cos ( 3 x ) - cos ( 4 x )}{x ^{2}})

x \to 0 lim (arctan (\frac{3}{x}))

x \to 0 - lim (x^{3} sin (\frac{π}{x}))

b \to 1 - lim (- ln (1 - b))

x \to a lim (\frac{1}{f ( x )})

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