What is the limit as x approaches 0 of (5^x-7^x)/x ?

The limit as x approaches 0 of (5^x-7^x)/x is ln(5)-ln(7)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

limit as x approaches 0 of (5^x-7^x)/x

x \to 0 lim (\frac{5 ^{x} - 7 ^{x}}{x})

ln (5) - ln (7)

Decimal Notation

- 0.33647 \dots

Solution steps

x \to 0 lim (\frac{5 ^{x} - 7 ^{x}}{x})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{5 ^{x} ln ( 5 ) - 7 ^{x} ln ( 7 )}{1})

= x \to 0 lim (\frac{5 ^{x} ln ( 5 ) - 7 ^{x} ln ( 7 )}{1})

Plug in the value

x = 0

= \frac{5 ^{0} ln ( 5 ) - 7 ^{0} ln ( 7 )}{1}

Simplify

\frac{5 ^{0} ln ( 5 ) - 7 ^{0} ln ( 7 )}{1} : ln (5) - ln (7)

= ln (5) - ln (7)

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

\frac{d y}{d x} (2) - 4 x y = 8 x

\int 8^{3 x + 1} d x

\frac{d}{d x} (\frac{x ^{2}}{x ^{2} - 25})

\int \frac{( 1 + y ^{2} )}{y} d y

What is the limit as x approaches 0 of (5^x-7^x)/x ?
The limit as x approaches 0 of (5^x-7^x)/x is ln(5)-ln(7)