Can anyone explain how line-$4$ follows from line-$3$? I don't know how line-$4$ follows from line-$3$.

 A: If $$L=\lim_{x\to a} f(a)^{g(x)}
\to 1^\infty \implies L=\exp[\lim_{x\to a} g(x)(f(x)-1)]$$
So $$L=\lim_{n\to \infty} \left(\frac{a^{1/n}+b^{1/n}+c^{1/n}}{3}\right)^{n}
=\lim_{n\to \infty} \exp[n\left(\frac{a^{1/n}+b^{1/n}+c^{1/n}}{3}-1\right). $$
$$\implies L=\exp\left(\frac{1}{3}\left[\lim_{m\to 0}\left(\frac{a^m-1}{m}\right)+\lim_{m\to 0}\left(\frac{b^m-1}{m}\right)+\lim_{m\to 0}\left(\frac{c^m-1}{m}\right)\right]\right).$$
$$\implies L=\exp[{\frac{1}{3}(\ln a+\ln b+\ln c})]=\exp(\ln (abc)^{1/3})=(abc)^{1/3}.$$
Here we have used $\lim_{x\to 0} \frac{a^x-1}{x}=\ln a.$
A: I'm guessing this is your issue:  Let's take the derivative of $f(x) = \ln \left( \frac{5^{1/x}}{7}\right).$
The derivative is the reciprocal of the argument of $\ln$ times the derivative of the argument:
$$f'(x) = \frac{7}{5^{1/x}} \frac{d}{dx}\left(\frac{5^{1/x}}{7}\right).$$
The $7$'s cancel:
$$f'(x) = 5^{-1/x} \frac{d}{dx}\left(5^{1/x}\right).$$
Using the rule $d/dx(a^x) = \ln(a) a^x$ and the chain rule we have:
$$f'(x) = 5^{-1/x}\ln(5) 5^{1/x}\frac{-1}{x^2}. $$
A: To go from Line 3 to Line 4, you use that as $n$ approaches infinity, $a^{1/n}$ approaches $a^0=1$ (and similarly for the other variables - as long as they're not zero), as well as log rules.
The $3/(a^{1/n}+b^{1/n}+c^{1/n})$ approaches $3/(1+1+1)=3/3=1$.
The $-a^{1/n}\ln(a)/3-b^{1/n}\ln(b)/3-c^{1/n}\ln(c)/3$ approaches $-(\ln(a)+\ln(b)+\ln(c))/3=-\ln(abc)/3$ using log rules.
Finally, the negative cancels with the negative in the denominator.
