What is the limit as p approaches 0 of (ln(0.7x^p+0.3y^p))/p ?

The limit as p approaches 0 of (ln(0.7x^p+0.3y^p))/p is 0.7ln(x)+0.3ln(y)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

limit as p approaches 0 of (ln(0.7x^p+0.3y^p))/p

p \to 0 lim (\frac{ln ( 0.7 x ^{p} + 0.3 y ^{p} )}{p})

0.7 ln (x) + 0.3 ln (y)

Solution steps

p \to 0 lim (\frac{ln ( 0.7 x ^{p} + 0.3 y ^{p} )}{p})

Factor out

0.7 x^{p}

from

0.7 x^{p} + 0.3 y^{p} : 0.7 x^{p} (1 + \frac{0.3 y ^{p}}{0.7 x ^{p}})

= p \to 0 lim \frac{ln ( 0.7 x ^{p} ( 1 + \frac{0.3 y ^{p}}{0.7 x ^{p}} ) )}{p}

Simplify

\frac{ln ( 0.7 x ^{p} ( 1 + \frac{0.3 y ^{p}}{0.7 x ^{p}} ) )}{p} : \frac{1}{p} ln (0.7 x^{p} + 0.3 y^{p})

= p \to 0 lim (\frac{1}{p} ln (0.7 x^{p} + 0.3 y^{p}))

Simplify

\frac{1}{p} ln (0.7 x^{p} + 0.3 y^{p}) : \frac{ln ( 0.7 x ^{p} + 0.3 y ^{p} )}{p}

= p \to 0 lim (\frac{ln ( 0.7 x ^{p} + 0.3 y ^{p} )}{p})

Apply L'Hopital's Rule

: p \to 0 lim (\frac{\frac{0.7 x ^{p} l n ( x ) + 0.3 y ^{p} l n ( y )}{0.7 x ^{p} + 0.3 y ^{p}}}{1})

= p \to 0 lim (\frac{\frac{0.7 x ^{p} l n ( x ) + 0.3 y ^{p} l n ( y )}{0.7 x ^{p} + 0.3 y ^{p}}}{1})

Plug in the value

p = 0

= \frac{\frac{0.7 x ^{0} l n ( x ) + 0.3 y ^{0} l n ( y )}{0.7 x ^{0} + 0.3 y ^{0}}}{1}

Simplify

\frac{\frac{0.7 x ^{0} l n ( x ) + 0.3 y ^{0} l n ( y )}{0.7 x ^{0} + 0.3 y ^{0}}}{1} : 0.7 ln (x) + 0.3 ln (y)

= 0.7 ln (x) + 0.3 ln (y)

x \to \infty lim (- 2 x + 11)

x \to 0 lim (\frac{sin ( x )}{0})

x \to 5 lim (e^{x})

x \to 5 lim (ln (x))

x \to 0 lim (cot (0))

What is the limit as p approaches 0 of (ln(0.7x^p+0.3y^p))/p ?
The limit as p approaches 0 of (ln(0.7x^p+0.3y^p))/p is 0.7ln(x)+0.3ln(y)