Evaluating a limit similar to the representation of $\frac{1}{e}$

Question

I want to evaluate the following limit:

$$ \lim_{x \to \infty} \left(1 - \frac{1}{b\cdot log_b(x)}\right)^{c \cdot (log_b(x))^2} $$

Where $b$ and $c$ are constants. Its form suggests to me that I should be able to use the fact that $\lim_{x \to \infty} \left(1 - \frac{1}{x}\right)^x = \frac{1}{e}$, but I'm having trouble figuring out how to proceed with it. Would appreciate any assistance.

Update:

Here is the direction I'm going in now. I can rewrite the limit as follows:

$$ \lim_{x \to \infty} \left(\left(1 - \frac{1}{b\cdot log_b(x)}\right)^{(log_b(x))^2}\left(1 - \frac{1}{b\cdot log_b(x)}\right)^{c}\right) $$

$$= \lim_{x \to \infty} \left(1 - \frac{1}{b\cdot log_b(x)}\right)^{(log_b(x))^2} \lim_{x \to \infty}\left(1 - \frac{1}{b\cdot log_b(x)}\right)^{c} $$ $$= \lim_{x \to \infty} \left(1 - \frac{1}{b\cdot log_b(x)}\right)^{(log_b(x))^2} 1^{c} $$ $$= \lim_{x \to \infty} \left(1 - \frac{1}{b\cdot log_b(x)}\right)^{(log_b(x))^2} $$

That gets rid of the the constant $c$, but I'm still unsure as to what to do given that $b$ occurs as a coefficient in the denominator but not in the exponent.

Update 2:

Here's the rest of the solution, reached with some nudging from Alex and Marty:

$$= \lim_{x \to \infty} \left(1 - \frac{1}{b\cdot log_b(x)}\right)^{(log_b(x))^2} $$
$$= \lim_{x \to \infty} \left(\left(1 - \frac{1}{b\cdot log_b(x)}\right)^{(b \cdot log_b(x))}\right)^{\frac{log_b(x)}{b}} $$
$$= \lim_{x \to \infty} \left(\frac{1}{e}\right)^{(log_b(x))^2} $$
$$= 0 $$

Answer 1

Remember that $u^{v^2} =u^{v\cdot v} =(u^v)^v $.

Therefore

$\begin{align*} \left(1 - \frac{1}{b\cdot log_b(x)}\right)^{c \cdot (log_b(x))^2} &= \left(\left(1 - \frac{1}{b\cdot log_b(x)}\right)^{c\,log_b(x)}\right)^{log_b(x)}\\ &= \left(\left(1 - \frac{1}{b\cdot log_b(x)}\right)^{b\,log_b(x)}\right)^{(c/b)log_b(x)}\\ &\to \left(1/e\right)^{(c/b)log_b(x)}\\ &\to 0\\ \end{align*} $