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.

  • 2
    $\begingroup$ $\lim_{x \to \infty} \left( 4^4+\frac{3^x}{4^x} \times 4^2 \right)=\lim_{x \to \infty}4^4 +\lim_{x \to \infty}\left(\dfrac{3^x}{4^x}\right)4^2$? While its true you might want to explain for rigor. $\endgroup$ Dec 22, 2018 at 4:26
  • $\begingroup$ 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. $\endgroup$
    – David K
    Jan 8, 2019 at 14:28

2 Answers 2


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.)


Yup, the answer is correct.


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

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