# Evaluating $\lim\limits_{x \to \infty}\frac{4^{x+2}+3^x}{4^{x-2}}$

$$\lim_{x→∞}\frac{4^{x+2}+3^x}{4^{x-2}}.$$

I have solved it like below: $$\lim_{x→∞}\left(\frac{4^{x+2}}{4^{x-2}}+\frac{3^x}{4^{x-2}}\right)=\lim_{x→∞}\left(4^4+\frac{3^x}{4^x}·4^2\right).$$ Since, as $$x → ∞$$, $$3^x → ∞$$, $$\dfrac{3^x}{4^x} → 0$$, the limit is equal to $$4^4=256$$.

Have I solved it correctly?

This was a practice test question and the given solution was wrong. So, I solved it and I am preparing alone, no friend to discuss, so I posted it here.

• $\lim_{x \to \infty} \left( 4^4+\frac{3^x}{4^x} \times 4^2 \right)=\lim_{x \to \infty}4^4 +\lim_{x \to \infty}\left(\dfrac{3^x}{4^x}\right)4^2$? While its true you might want to explain for rigor. – Yadati Kiran Dec 22 '18 at 4:26
• It's unclear to me what purpose it serves to write $3^x\to\infty$. It makes the next expression, $\frac{3^x}{4^x} \to 0$, look wrong even though it actually is correct. – David K Jan 8 at 14:28

Yup! Your steps almost all follow logically and evaluate to the correct limit, so your solution is correct (almost).

I will make a nitpick with one thing you said, though:

Since, as $$x \to \infty, 3^x \to \infty, \frac{3^x}{4^x} \to 0$$

Technically, since $$3^x$$ and $$4^x$$ both approach $$\infty$$ as $$x \to \infty$$, then under your logic we obtain an indeterminate form:

$$\frac{3^x}{4^x} \to \frac{\infty}{\infty}$$

It would be better to regroup $$3^x/4^x$$ as $$(3/4)^x$$. Then clearly, as $$x \to \infty$$, $$(3/4)^x \to 0$$ because $$3/4 < 1$$. (If you're not convinced, notice if you take $$x = 1, 2, 3, 4$$ and so on that $$(3/4)^x$$ clearly is decreasing.)

• While $$3^x \to \infty$$ is true, note that it is not used in the proof though.