Evaluate the following limits: _ x 1 [a b^x-a^x b x^2-1 ]

Question

Evaluate the following limits:

$
\lim _{x \rightarrow 1}\left[\frac{a b^x-a^x b}{x^2-1}\right]
$

✓

Answer

$
\begin{aligned}
\lim _{x \rightarrow 1} \frac{a b^x-\mathrm{a}^x \mathrm{~b}}{x^2-1} & =\lim _{x \rightarrow 1} \frac{\mathrm{ab}\left(\mathrm{b}^{x-1}-\mathrm{a}^{x-1}\right)}{x^2-1^2} \\
& =\lim _{x \rightarrow 1} \frac{\mathrm{ab}\left(\mathrm{b}^{x-1}-\mathrm{a}^{x-1}\right)}{(x-1)(x+1)}
\end{aligned}
$
Put $x=1+\mathrm{h}$,
$\therefore \quad x-1=\mathrm{h}$
As $x \rightarrow 1, \mathrm{~h} \rightarrow 0$
$\therefore \quad \lim _{x \rightarrow 1} \frac{a b^x-a^x b}{x^2-1}=\lim _{h \rightarrow 0} \frac{a b\left(b^h-a^h\right)}{h(1+h+1)}$
$=a b \lim _{h \rightarrow 0} \frac{b^h-1+1-a^h}{h(2+h)}$
$=a b \lim _{h \rightarrow 0} \frac{\left(b^h-1\right)-\left(a^h-1\right)}{h(2+h)}$
$=a b \lim _{h \rightarrow 0} \frac{1}{2+h}\left(\frac{b^h-1}{h}-\frac{a^h-1}{h}\right)$
$=a b \cdot \frac{1}{\lim _{h \rightarrow 0}(2+h)}\left(\lim _{h \rightarrow 0} \frac{b^h-1}{h}-\lim _{h \rightarrow 0} \frac{a^h-1}{h}\right)$
$=$ ab. $\frac{1}{2+0} \cdot(\log \mathrm{b}-\log \mathrm{a}) \cdots\left[\because \lim _{x \rightarrow 0} \frac{\mathrm{a}^x-1}{x}=\log \mathrm{a}\right]$
$=\frac{a b}{2} \log \left(\frac{b}{a}\right)$

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