Question

Suppose that $a$ and $b(b\ne a)$ are real positive numbers, then the value of $\underset{t\to 0}{lim}{\left(\frac{b^{t+1}-a^{t+1}}{b-a}\right)}^{1/t}$ is equal to

• A. $\frac{a\ln b-b\ln a}{b-a}$
• B. $\frac{b\ln b-a\ln a}{b-a}$
• C. $b\ln b-a\ln a$
• D. ${\left(\frac{b^b}{a^a}\right)}^{\frac{1}{b-a}}$

Answer 1

The correct option is D ${\left(\frac{b^b}{a^a}\right)}^{\frac{1}{b-a}}$

$\underset{t\to 0}{lim}{\left(\frac{b^{t+1}-a^{t+1}}{b-a}\right)}^{1/t},1^{\infty }$ form

$=e^{\underset{t\to 0}{lim}\frac{1}{t}[\frac{b^{t+1}-a^{t+1}}{b-a}-1]}$

$=e^{\underset{t\to 0}{lim}\left(\frac{b^{t+1}-a^{t+1}-b+a}{t(b-a)}\right)}$

$=e^{\underset{t\to 0}{lim}\left(\frac{b(b^t-1)-a(a^t-1)}{t(b-a)}\right)}$

$=e^{\underset{t\to 0}{lim}\left(\frac{b\left(\frac{b^t-1}{t}\right)-a\left(\frac{a^t-1}{t}\right)}{(b-a)}\right)}$

$\because \underset{t\to 0}{lim}\frac{x^t-1}{t}=\ln x$

$=e^{\frac{b\ln b-a\ln a}{b-a}}$
$=e^{\frac{\ln b^b-\ln a^a}{b-a}}$
$=e^{\frac{\ln \frac{b^b}{a^a}}{b-a}}$
$=e^{\ln {\left(\frac{b^b}{a^a}\right)}^{\frac{1}{b-a}}}={\left(\frac{b^b}{a^a}\right)}^{\frac{1}{b-a}}$