What is the limit as t approaches 0 of (8^t-5^t)/t ?

The limit as t approaches 0 of (8^t-5^t)/t is 3ln(2)-ln(5)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

limit as t approaches 0 of (8^t-5^t)/t

t \to 0 lim (\frac{8 ^{t} - 5 ^{t}}{t})

3 ln (2) - ln (5)

Decimal Notation

0.47000 \dots

Solution steps

t \to 0 lim (\frac{8 ^{t} - 5 ^{t}}{t})

Apply L'Hopital's Rule

: t \to 0 lim (\frac{3 ln ( 2 ) \cdot 8 ^{t} - 5 ^{t} ln ( 5 )}{1})

= t \to 0 lim (\frac{3 ln ( 2 ) \cdot 8 ^{t} - 5 ^{t} ln ( 5 )}{1})

Plug in the value

t = 0

= \frac{3 ln ( 2 ) \cdot 8 ^{0} - 5 ^{0} ln ( 5 )}{1}

Simplify

\frac{3 ln ( 2 ) \cdot 8 ^{0} - 5 ^{0} ln ( 5 )}{1} : 3 ln (2) - ln (5)

= 3 ln (2) - ln (5)

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

\frac{\partial}{\partial u} (v sin (u))

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

\frac{d}{d x} (\frac{3 - 8 x}{x + 4})

What is the limit as t approaches 0 of (8^t-5^t)/t ?
The limit as t approaches 0 of (8^t-5^t)/t is 3ln(2)-ln(5)