Limit what do i "pull" in front of the brackets first

Question

I am in a doubt with this simple problem. What is the correct answer to this problem: $\lim_{x\to-∞}{\frac{x+3^x}{3x-3}}$

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

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

I know that $3^x$ "grows" faster so I think $(1)$ would be correct?

Answer 1

Let $y:=-x.$

Then consider :

$\lim_{y \rightarrow \infty}\dfrac{-y+3^{-y}}{-3y-3}.$

$\dfrac{-y+3^{-y}}{-3y-3} =(1/3)\dfrac{y-3^{-y}}{y+1}=$

$(1/3)\dfrac{1-3^{-y}/y}{1+1/y} .$

The limit $y \rightarrow \infty$ is?

Note: Your expression 2: You need to consider $3^x$ and $x \rightarrow - \infty$.

The above is similar to your version 2) , find it easier to deal with $y \rightarrow + \infty.$

Answer 2

$3^x$ doesn't grow as $x \to -\infty$; instead, it converges (very quickly) to zero.

If you try computing the limit via formula 1, after cancelling its of the form $((-\infty) + 1)/((-\infty) - \infty)$; not a form suitable for directly computing the limit, and with no obvious leads for further manipulation (other than undoing the previous step).

The only significant terms are the $x$ in the numerator and the $3x$ in the denominator; your formula 2 is a an effective way to isolate the significant part so that it can be cancelled out.

Answer 3

No $(2)$ is correct; because $3^x\to 0$ as $x\to -\infty$.

However, it's so simpler with equivalents:

We have $x+3^x\sim_{-\infty} x$ so $$\frac{x+3^x}{3x-3}\sim_{-\infty}\frac{x}{3x}=\frac13.$$

Answer 4

Simply note that for $x\to-\infty$ we have $3^x\to0$ thus the leading terms are $x$ and $3x$.