# Evaluate $\lim_{x\to \infty}\frac{x^{2018}+2017^x}{x^{2017}+2018^x}$

I was posed this problem in my test today. $$\lim_{x\to \infty}\frac{x^{2018}+2017^x}{x^{2017}+2018^x}$$

My Attempt:

I saw that $x^{2018}$ and $x^{2017}$ were vanishing on derivating so,

I applied L Hospital on this $2019$ times, and it reduced to something like this,

$$\lim_{x\to \infty}\bigg( \frac{2017}{2018}\bigg)^x\bigg( \frac{\ln{2017}}{\ln{2018}}\bigg)^{2019}$$ as, $\frac{2017}{2018}\lt 1$

this limit will coverge to zero.

Am I solving this right? Also this method seems a little brute, what other approach can I take to this question?

• You must have a lot of time on test to apply it $2019$ times ;). Yes, this method is ok. $\lim\limits_{x\to \infty} x^ka^x = 0$ for any $0<a<1$ and any $k$ is a known result. You could of just divided by $2018^x$. But this approach is safer, because you never know if you can use some theorems – Jakobian Jun 2 '18 at 13:42
• @Adam I just saw that $x^k$ will vanish after $k+1$ derivates and directly took the $2019^{th}$derivative. – prog_SAHIL Jun 2 '18 at 13:52

You are correct that $x^{2017}$ and $x^{2018}$ don't matter when $x$ gets large enough. If you ignore them you have $$\lim_{x \to \infty}\left(\frac {2017}{2018}\right)^x$$ As the fraction is less than $1$ the limit is $0$. To be a little more formal about it, divide numerator and denominator by $2018^x$ and note that the denominator is greater than $1$ and the numerator goes to zero.