Why is the limit of this expression not zero?

Question

$$\lim_{n\to \infty} {-7^n + 8^{n-2}\over 7^{n+1} + 8^{n+2}} = \frac {1}{4096}$$

I don’t understand why this is the limit. Since the denominator has an $8^{n+2}$, isn’t the limit supposed to be 0? When you have a quotient of polynomials, and the denominator has the greatest exponent, isn’t the limit supposed to be 0?

Any help is appreciated.

PS: The fraction is supposed to be the general term of a series, so L’Hôpital’s rule does not apply.

Answer 1

HINT:

$$\frac{-7^n + 8^{n-2}}{7^{n+1} + 8^{n+2}} =\frac{-(7/8)^n+8^{-2}}{7(7/8)^n+8^2}$$

Now as $-1<7/8<1,\lim_{n\to\infty}(7/8)^n=0$

Answer 2

Hint: write $$ {-7^n + 8^{n-2}\over 7^{n+1} + 8^{n+2}} = {-\left(\frac78\right)^n + \frac 1{64} \over 7 \left(\frac78\right)^n + 64} $$

Answer 3

As often it's shorter with equivalents: as $7^n=o(8^{n-2})$ since $7<8$, $$-7^n+8^{n-2}\sim_{\infty}8^{n-2}$$ Similarly, $\,7^{n+1}+8^{n+2}\sim_{\infty}8^{n+2}$, hence: $$\frac{-7^n+8^{n-2}}{7^{n+1}+8^{n+2}}\sim_{\infty}\frac{8^{n-2}}{8^{n+2}}=\frac 1{8^4}=\frac1{4096}.$$