Simplifying $ \lim x\rightarrow\infty \frac {(1+2x^{1/6})^{2016}}{1+(2+(3+4x^6)^7)^8}$

Question

So I have a problem regarding limits in my calculus class:

$$ \lim x\rightarrow\infty \frac {(1+2x^{1/6})^{2016}}{1+(2+(3+4x^6)^7)^8}$$

Basically what I've identified is that it's an $\frac{\infty}{\infty}$ expression and thus tried applying l'Hôpital.

Doing this I've gotten from there, to here:

$$\lim x\rightarrow\infty \frac{x^{-5/6}(1+2x^{1/6})^{2015}}{2x^{5}(2+(3+4x^{6})^7)^{7}(3+7x^{6})^{6}}$$

So what I've come to realize is that applying l'H again won't simplify my problem (I tried and it wasn't pretty). As I see the equation, as $x\rightarrow\infty$ the answer has to be $\infty$. My problem is basically how I simplify the expression (if it is possible) as I'm guessing I need to prove it in some way.

I've already seen this post with the same problem: Help solving a limit problem with infinity and large exponents

But I'm really struggling to grasp the rewrite done as the solution.

Any help would be greatly appreciated.

Thanks in advance.

Answer 1

One may write, as $x \to \infty$, $$ \begin{align} \frac {(1+2x^{1/6})^{2016}}{1+(2+(3+4x^6)^7)^8}&= \frac {\left(1+\frac1{2x^{1/6}}\right)^{2016}\times2^{2016}x^{336}}{1+\left(2+\left(1+\frac3{4x^6}\right)^7\times4^7x^{42}\right)^8} \\\\&= \frac {\left(1+\dfrac1{2x^{1/6}}\right)^{2016}\times2^{2016}\color{red}{x^{336}}}{\left(\dfrac2{4^{56}x^{336}}+\left(\dfrac2{4^7x^{42}}+\left(1+\dfrac3{4x^6}\right)^7\right)^8\right)\times4^{56}\color{red}{x^{336}}} \\\\&\to\frac{2^{2016}}{4^{56}}=\color{blue}{2^{1904}}. \end{align} $$

Answer 2

Note: Use \to instead of \rightarrow for limits.

First of all, the degree(Largest exponent) of the numerator is $336$, and the degree of the denominator is $336$. If they differ then the result follows immediately, but in this case they are equal. It is safe to throw away all terms with smaller exponents(divide by the largest term to see this), leaving you $$ \lim_{x\to\infty} \frac {(1+2x^{1/6})^{2016}}{1+(2+(3+4x^6)^7)^8} = \lim_{x\to\infty} \frac {(2x^{1/6})^{2016}}{((4x^6)^7)^8} $$ Then cancel out all $x$es and conclude the limit is...

Answer 3

Use the fact that for any two polynomials of same degree $P(x), Q(x)$, $\lim_{x\to\infty} \frac{P(x)}{Q(x)}$ is equal to ratio of their leading coefficients.

Let $P(x)=a_nx^n+a_{n-1}x^{n-1}+\ldots+a_0$, $Q(x)=b_nx^n+b_{n-1}x^{n-1}+\ldots+b_0$. Then, $$\frac{P(x)}{Q(x)}=\frac {x^n(a_n+\overbrace{a_{n-1}x^{-1}+\ldots+a_0x^{-n}}^{\text{goes to 0 as }n\to\infty})} {x^n(b_n+\underbrace{b_{n-1}x^{-1}+\ldots+b_0x^{-n}}_{\text{goes to 0 as }n\to\infty})}=\frac{a_n}{b_n}$$

As both polynomials have degree $2016/6=6\times 7\times8=366$, the limit is $\frac{2^{2016}}{4^{366}}=2^{1344}$