Limit Question: doubting about my answer! A limit where $x \to \infty$

Question

Hey guys so I have this limit:

$$\lim_{x \to ∞} f(x) = \frac{(3x+1)^3(x-1)}{(x-2)^4}$$

and I got $27$ as final answer, just wondering if you guys can check; I expanded the numerator and denominator and then divided everything by $x^4$

Answer 1

Rather than expanding, divide top and bottom by $x^4$ directly.

$$\lim_{x \to \infty}\frac{(3x+1)^3(x-1)}{(x-2)^4}=\lim_{x \to \infty}\frac{(3+\frac1x)^3(1-\frac1x)}{(1-\frac2x)^4}=\frac{27(1)}1=27$$

$27$ is the correct answer.

Answer 2

Note that

$$\frac{(3x+1)^3(x-1)}{(x-2)^4}=27\frac{(x+1/3)^3(x-1)}{(x-2)^4}=27\frac{\frac{(x+1/3)^3}{x^3}\frac{(x-1)}x}{\frac{(x-2)^4}{x^4}}\\=27\frac{(1+1/x)^3(1-1/x)}{(1-2/x)^4}\to27\frac{(1+0)^3(1-0)}{(1-0)^4}=27$$

as $x\to\infty$, so your result is correct.

Answer 3

Yes, your answer is correct. An easy way is

\begin{align} \lim_{x \to ∞} f(x) &=\lim_{x \to ∞} \frac{(3x+1)^3(x-1)}{(x-2)^4}\\ &= \lim_{x \to ∞}\frac{x^4\cdot (3+\frac{1}{x})^3(1-\frac{1}{x})}{x^4\cdot (1-\frac{2}{x})^4}\\ &= \lim_{x \to ∞} \frac{(3+\frac{1}{x})^3(1-\frac{1}{x})}{ (1-\frac{2}{x})^4}\\ &=27 \end{align}

Answer 4

You don't need to really check too much. It's sufficient to see that:

$$(3x+1)^3(x-1) = 3^3x^4 + \text{(other stuff of a lesser degree)}$$

and

$$(x-2)^4 = x^4 + \text{(other stuff of a lesser degree)}$$

Since the numerator and denominator have the same degree, the limit as $x\to \infty$ will be the ratio of the highest-degree terms' coefficients. Since the coefficients of the $x^4$ terms, respectively, are $3^3 = 27$ and $1$, then the limit evaluates to

$$\lim_{x \to ∞} f(x) = \frac{(3x+1)^3(x-1)}{(x-2)^4} = \frac{27}{1} = 27$$

So, tl;dr version, you're right. The argument could be made that you did a bit more work than necessary by multiplying everything out, but that's more an issue of personal taste in a case like this (since it's not a ton more work).

Answer 5

Only to check or guess the possible result we can use that the constant term are negligeble then

$$\frac{(3x+1)^3(x-1)}{(x-2)^4}\sim \frac{(3x)^3(x)}{(x)^4}=\frac{27x^4}{x^4}=27$$

and then proceed for a rigorous proof as already indicated.

Answer 6

As an alternative method (for your practicing) you can rearrange degrees: $$\lim_{x \to ∞} \frac{(3x+1)^3(x-1)}{(x-2)^4}=\lim_{x \to ∞} \left(\frac{3x+1}{x-2}\right)^3\left(\frac{x-1}{x-2}\right)=\cdots=3^3\cdot 1=27.$$ Note: In the dots you should use the properies of limits.