Question

Suppose that $a$ and $b$ $(b\ne a)$ are real positive number the value of $\underset{t\to 0}{lim}(\frac{b^{1+t}-a^{1+t}}{b-a}{)}^{1/t}$ has the value equals 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. $(\frac{b^b}{a^a}{)}^{\left(1/\right(b-a\left)\right)}$

Answer 1

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

Given ${lim}_{t\to 0}{\left(\frac{b^{1+t}-a^{1+t}}{b-a}\right)}^{\frac{1}{t}}a\ne b$

If $L=\underset{x\to a}{lim}{f\left(x\right)}^{g\left(x\right)}$

$\underset{x\to a}{lim}f\left(x\right)\to 1and\underset{x\to a}{lim}g\left(x\right)\to \infty$

$\Rightarrow L=e^{\underset{x\to a}{lim}(f\left(x\right)-1).g\left(x\right)}$
Thus,
$L=\underset{t\to 0}{lim}{e}^{(\frac{b^{1+t}-a^{1+t}}{b-a}-1).\frac{1}{t}}$

$=e^{\frac{1}{(b-a)}\left[{lim}_{t\to 0}\frac{1}{t}\{b^{t+1}-a^{t+1}\}-(b-a)\right]}\left(\frac{0}{0}\right)$ form
Using L'Hospital's Rule

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

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

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