What is the limit as x approaches infinity of ((ln(3x)))/(e^{7x)} ?

The limit as x approaches infinity of ((ln(3x)))/(e^{7x)} is 0

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

limit as x approaches infinity of ((ln(3x)))/(e^{7x)}

x \to \infty lim (\frac{( ln ( 3 x ) )}{e ^{7 x}})

0

Solution steps

x \to \infty lim (\frac{ln ( 3 x )}{e ^{7 x}})

Apply L'Hopital's Rule

: x \to \infty lim (\frac{\frac{1}{x}}{e ^{7 x} \cdot 7})

= x \to \infty lim (\frac{\frac{1}{x}}{e ^{7 x} \cdot 7})

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= x \to \infty lim (\frac{1}{7 e ^{7 x} x})

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

= \frac{1}{7} \cdot x \to \infty lim (\frac{1}{e ^{7 x} 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{1}{7} \cdot \frac{lim _{x \to \infty} ( 1 )}{lim _{x \to \infty} ( e ^{7 x} x )}

x \to \infty lim (1) = 1

x \to \infty lim (e^{7 x} x) = \infty

= \frac{1}{7} \cdot \frac{1}{\infty}

Simplify

\frac{1}{7} \cdot \frac{1}{\infty} : 0

= 0

n = 0 \sum \infty (- 5)^{n}

\int_{0}^{1} (1 + 4 x y) d x

y = 3 + 4 x^{2} - 2 x^{3}, (2, 3)

\frac{d}{d x} (\frac{4 + csc ( x )}{8 - csc ( x )})

\int 4 x^{3} ln (x) d x

What is the limit as x approaches infinity of ((ln(3x)))/(e^{7x)} ?
The limit as x approaches infinity of ((ln(3x)))/(e^{7x)} is 0