Suppose given that
$$ \lim_{x \to 0}f(x)=1 \\ \lim_{x \to 0}g(x)=\infty \\ \lim_{x \to 0}g(x)(f(x)-1)=c $$
Then find the value of $ \lim_{x \to 0} f(x)^{g(x)} $
So this is what I've done. Please verify if its correct approach @Macavity $$ f(x)^{g(x)} = e^{\log{f(x)^{g(x)}}} = e^{{g(x)\log f(x)}} $$ $(1)$
Now from Taylor series expansion of $\log(1+x)$ Substitute $x$ by $f(x)-1 $ in the Taylor series $$\log f(x)= (f(x)-1) - \frac12(f(x)-1)^2 \cdots$$ $(2)$
From $(1)$ and $(2)$ By taking only the first term of the $\log f(x) $ expansion
$$ \lim_{x \to 0} f(x)^{g(x)}= e^c $$
Is this approach correct?
Your approach is intuitively correct. But a simpler approach with rigorous justification of all the steps is as follows.
Let $L = \lim_{x \to 0}f(x)^{g(x)}$ so that \begin{align}\log L &= \log\left(\lim_{x \to 0}f(x)^{g(x)}\right)\notag\\ &=\lim_{x \to 0}\log f(x)^{g(x)}\text{ (by continuity of log)}\notag\\ &= \lim_{x \to 0}g(x)\log f(x)\notag\\ &= \lim_{x \to 0}g(x)\log \{1 + f(x) - 1\}\notag\\ &= \lim_{x \to 0}g(x)(f(x) - 1)\dfrac{\log \{1 + f(x) - 1\}}{f(x) - 1}\notag\\ &= \lim_{x \to 0}g(x)(f(x) - 1)\cdot\lim_{t \to 0}\dfrac{\log (1 + t)}{t}\text{ (by putting }t = f(x) - 1)\notag\\ &= c\cdot 1 = c\end{align} and hence $L = e^{c}$. We have used the standard limit $$\lim_{t \to 0}\frac{\log(1 + t)}{t} = 1$$ Also note that the second condition $\lim_{x \to 0}g(x) = \infty$ is unnecessary as can be seen in the above derivation.