# Calculating a limit with exponents

There is a limit I want to calculate, and I've calculated it as follows:

$$\lim_{x \to \infty} \frac{7^{-x+1}-2\cdot 5^{-x}}{3^{-x}-7^{-x}} = \lim_{x \to \infty} \frac{\frac{7}{7^x}-\frac{2}{5^x}}{\frac{1}{3^x}-\frac{1}{7^x}}$$

If I now multiply the denominator and numerator by $$3^x$$, I get the following:

$$\lim_{x \to \infty} \frac{7\cdot\frac{3^x}{7^x}-2\cdot\frac{3^x}{5^x}}{1-\frac{3^x}{7^x}} = \frac{0}{1}.$$

Is it correct to assume that the fractions in the numerator will limit to $$0$$ and not to infinity (because they are multiplied by $$7$$ and $$2$$ respectively)?

Does this way of thinking apply to all limits of such format (except for those that limit to $$e$$) or are there exceptions?

• As $7<5,3$, the terms in $7^{-x}$ become negligible and the expression is asymptotic to $-2\left(\dfrac35\right)^{x}$.
– user1015917
Jan 23 at 14:56
• But I would say that "becomes negligible" is a step beyond OP's analysis. Jan 23 at 22:55

Your answer is correct. In order to be sure it's nice to write : $$\lim_{x \to +\infty} \dfrac{7 \dfrac{3^x}{7^x} - 2 \dfrac{3^x}{5^x}}{1 - \dfrac{3^x}{7^x}} = \dfrac{7 \times 0 - 2 \times 0}{1 - 0} = \dfrac{0}{1} = 0$$