Evaluation of the limit $\lim_{q\rightarrow 1} \frac{\phi^5(q)_{\infty}}{\phi(q^5)_{\infty}}$ Given the Euler function $\phi(q)=\prod_{n = 1}^{\infty}(1-q^{n})$ which is a modular form where $q=\exp(2\pi i \tau)$, $|q|\lt1$

Then what is the limit $\lim_{q\rightarrow 1}\frac{\phi^5(q)_{\infty}}{\phi(q^5)_{\infty}}$ ?


Given the following identity $\frac{(q)^5_{\infty}}{(q^5)_{\infty}}=1+5\sum_{n=1}^{\infty}\frac{(-1)^n nq^{\frac{n(n+1)}{2}}}{\prod_{k=1}^{n-1}\Big(R_{k}+q^k\Big)}$ from my old post where $\frac{1}{R_{n}}=R(q^n,q)$ is the Generalized Rogers Ramanujan continued fraction
If we consider the fact that $\lim_{x\rightarrow 0+}\frac{1}{R(e^{-nx},e^{-x})}=\Phi$ 
where $\Phi=\frac{1+\sqrt5}{2}$ is the golden ratio 

Then we are led to $$\lim_{q\rightarrow 1}\frac{\phi^5(q)_{\infty}}{\phi(q^5)_{\infty}}=1+5\sum_{n=1}^{\infty}\frac{(-1)^n n}{(1+\Phi)^n}$$

Whereby the infinite series on the RHS converges to zero.
