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

The limit as x approaches infinity of (((-4)^{x+1}))/((x+1)*4^{x+1)} is 0

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

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

0

Solution steps

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

Apply exponent rule:

(a \cdot b)^{n} = a^{n} \cdot b^{n}

(- 4)^{x + 1} = (- 1)^{x + 1} (4)^{x + 1}

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

Cancel the common factor:

4^{x + 1}

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

Simplify

(- 1)^{x + 1} : (- 1)^{x} (- 1)

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

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

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

Divide by highest denominator power:

\frac{\frac{( - 1 ) ^{x}}{x}}{1 + \frac{1}{x}}

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

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

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

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

= - \frac{0}{1}

Apply rule

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

= 0

\frac{d}{d x} (4 sec (\frac{x}{2}))

\int sec^{2} (x) - 8 cot (x) d x

y^{^{''}} + 36 y = 24 sin (6 t)

y = \frac{1}{7 t ^{3}}

x \to 1 + lim ((1 - x)^{l n (x)})

What is the limit as x approaches infinity of (((-4)^{x+1}))/((x+1)*4^{x+1)} ?
The limit as x approaches infinity of (((-4)^{x+1}))/((x+1)*4^{x+1)} is 0