A difficult Limit problem consisting $n^{1/n}$ in a high school level exam . I was asked in an exam to solve the following question--

If $\lim_{n \to \infty}(a \sqrt[n]{n} + b)^{n/\ln n}$ has the
  value equal to $e^{-3}$ ,then find the value of $(4b+3a)$.

I tried with L'Hospital  I tried bringing the $\dfrac{n}{\ln n}$ down but nothing seems to work .
 A: Clearly on taking logs we have the following equation $$\lim_{n \to \infty}\frac{n}{\log n}\log(a\sqrt[n]{n} + b) = -3$$ from which we easily get $$\lim_{n \to \infty}\log(a\sqrt[n]{n} + b) = \lim_{n \to \infty}\frac{\log n}{n}\cdot\frac{n}{\log n}\log(a\sqrt[n]{n} + b) = 0 \cdot (-3) = 0$$ and hence $$\lim_{n \to \infty}(a\sqrt[n]{n} + b) = 1$$ which means $a + b = 1$. We also need to know that $a \neq 0$ because if it were so we would get $b = 1$ and hence $a\sqrt[n]{n} + b = 1$ and hence the limit in question would be $1$ instead of $e^{-3}$.
Next we need another equation between $a, b$. To that end we proceed as follows
\begin{align}
-3 &= \lim_{n \to \infty}\frac{n}{\log n}\log(a\sqrt[n]{n} + b)\notag\\
&= \lim_{n \to \infty}\frac{n}{\log n}\log(a + b + a\sqrt[n]{n} - a)\notag\\
&= \lim_{n \to \infty}\frac{n}{\log n}\log(1 + a\sqrt[n]{n} - a)\notag\\
&= \lim_{n \to \infty}\frac{n}{\log n}\cdot a(\sqrt[n]{n} - 1)\cdot\frac{\log(1 + a(\sqrt[n]{n} - 1))}{a(\sqrt[n]{n} - 1)}\text{ (here we use }a \neq 0)\notag\\
&= a\lim_{n \to \infty}\frac{n}{\log n}(\sqrt[n]{n} - 1)\notag\\
&= a\lim_{n \to \infty}\frac{n}{\log n}(\exp((\log n)/n) - 1)\notag\\
&= a\lim_{t \to 0}\frac{\exp(t) - 1}{t}\text{ (putting }t = (\log n)/n)\notag\\
&= a\notag
\end{align}
So we have $a = -3, b = 4$ and $4b + 3a = 7$.
Note: In the above we have used a famous limit $n^{1/n} \to 1$ which implies that $(\log n)/n \to 0$ and $a(\sqrt[n]{n} - 1) \to 0$. Further we have used the following limits $$\lim_{x \to 0}\frac{\log(1 + x)}{x} = 1,\,\lim_{x \to 0}\frac{e^{x} - 1}{x} = 1$$
