What is the limit as x approaches 0 of ((2^x-7^x))/x ?

The limit as x approaches 0 of ((2^x-7^x))/x is ln(2)-ln(7)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

limit as x approaches 0 of ((2^x-7^x))/x

x \to 0 lim (\frac{( 2 ^{x} - 7 ^{x} )}{x})

ln (2) - ln (7)

Decimal

- 1.25276 \dots

Solution steps

x \to 0 lim (\frac{2 ^{x} - 7 ^{x}}{x})

Apply L'Hopital's Rule

= x \to 0 lim (\frac{2 ^{x} ln ( 2 ) - 7 ^{x} ln ( 7 )}{1})

Plug in the value

x = 0

= \frac{2 ^{0} ln ( 2 ) - 7 ^{0} ln ( 7 )}{1}

Simplify

\frac{2 ^{0} ln ( 2 ) - 7 ^{0} ln ( 7 )}{1} : ln (2) - ln (7)

= ln (2) - ln (7)

What is the limit as x approaches 0 of ((2^x-7^x))/x ?
The limit as x approaches 0 of ((2^x-7^x))/x is ln(2)-ln(7)