Question

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Answer 1

On taking $\sf{x}$ directly as tending to infinity,

$\displaystyle\longrightarrow\sf{\lim_{x\to\infty}\dfrac{\log(x-c)}{\log(e^x-e^c)}=\dfrac{\log(\infty-c)}{\log(e^{\infty}-e^c)}}$

For $\sf{c}$ being smaller,

$\displaystyle\longrightarrow\sf{\lim_{x\to\infty}\dfrac{\log(x-c)}{\log(e^x-e^c)}=\dfrac{\log(\infty)}{\log(e^{\infty})}}$

$\displaystyle\longrightarrow\sf{\lim_{x\to\infty}\dfrac{\log(x-c)}{\log(e^x-e^c)}=\dfrac{\infty}{\infty}}$

So we have to make use of L'hospital's Rule.

$\displaystyle\longrightarrow\sf{\lim_{x\to a}\dfrac{f(x)}{g(x)}=\lim_{x\to a}\dfrac{f'(x)}{g'(x)}\quad\iff\quad\lim_{x\to a}\dfrac{f(x)}{g(x)}=\dfrac{0}{0}\quad OR\quad\lim_{x\to a}\dfrac{f(x)}{g(x)}=\dfrac{\pm\infty}{\pm\infty}}$

So,

$\longrightarrow\sf{\dfrac{d}{dx}\big[\log(x-c)\big]=\dfrac{1}{x-c}}$

And,

$\longrightarrow\sf{\dfrac{d}{dx}\big[\log(e^x-e^c)\big]=\dfrac{e^x}{e^x-e^c}}$

Then, by L'hospital's Rule,

$\displaystyle\longrightarrow\sf{\lim_{x\to\infty}\dfrac{\log(x-c)}{\log(e^x-e^c)}=\lim_{x\to\infty}\dfrac{\left(\dfrac{1}{x-c}\right)}{\left(\dfrac{e^x}{e^x-e^c}\right)}}$

$\displaystyle\longrightarrow\sf{\lim_{x\to\infty}\dfrac{\log(x-c)}{\log(e^x-e^c)}=\dfrac{\displaystyle\lim_{x\to\infty}\dfrac{1}{x-c}}{\displaystyle\lim_{x\to\infty}\dfrac{e^x}{e^x-e^c}}\quad\quad\dots(1)}$

We see that,

$\displaystyle\longrightarrow\sf{\lim_{x\to\infty}\dfrac{1}{x-c}=\dfrac{1}{\infty-c}}$

$\displaystyle\longrightarrow\sf{\lim_{x\to\infty}\dfrac{1}{x-c}=\dfrac{1}{\infty}}$

$\displaystyle\longrightarrow\sf{\lim_{x\to\infty}\dfrac{1}{x-c}=0}$

And,

$\displaystyle\longrightarrow\sf{\lim_{x\to\infty}\dfrac{e^x}{e^x-e^c}=\dfrac{e^{\infty}}{e^{\infty}-e^c}}$

$\displaystyle\longrightarrow\sf{\lim_{x\to\infty}\dfrac{e^x}{e^x-e^c}=\dfrac{e^{\infty}}{e^{\infty}}}$

$\displaystyle\longrightarrow\sf{\lim_{x\to\infty}\dfrac{e^x}{e^x-e^c}=\dfrac{\infty}{\infty}}$

Applying L'hospital's Rule,

$\displaystyle\longrightarrow\sf{\lim_{x\to\infty}\dfrac{e^x}{e^x-e^c}=\lim_{x\to\infty}\dfrac{\frac{d}{dx}(e^x)}{\frac{d}{dx}(e^x-e^c)}}$

$\displaystyle\longrightarrow\sf{\lim_{x\to\infty}\dfrac{e^x}{e^x-e^c}=\lim_{x\to\infty}\dfrac{e^x}{e^x}}$

$\displaystyle\longrightarrow\sf{\lim_{x\to\infty}\dfrac{e^x}{e^x-e^c}=\lim_{x\to\infty}1}$

$\displaystyle\longrightarrow\sf{\lim_{x\to\infty}\dfrac{e^x}{e^x-e^c}=1}$

Hence (1) becomes,

$\displaystyle\longrightarrow\sf{\lim_{x\to\infty}\dfrac{\log(x-c)}{\log(e^x-e^c)}=\dfrac{0}{1}}$

$\displaystyle\longrightarrow\sf{\underline{\underline{\lim_{x\to\infty}\dfrac{\log(x-c)}{\log(e^x-e^c)}=0}}}$

Hence 0 is the answer.