Limit of a function with multiple radicals at infinity

Question

$$ \lim_{x\to\infty} \frac{2x^{1/2}+3x^{1/3}+5x^{1/5}}{(3x-2)^{1/2}+(2x-3)^{1/3}} $$

I tried to divide the expresion with the highest power of $x$, but the problem is with the denominator.

Answer 1

$$ \lim_{x\to\infty} \frac{2x^{1/2}+3x^{1/3}+5x^{1/5}}{(3x-2)^{1/2}+(2x-3)^{1/3}} =\lim_{x\to\infty} \frac{2+3x^{1/3-1/2}+5x^{1/5-1/2}}{(3-2/x)^{1/2}+x^{1/3-1/2}(2-3/x)^{1/3}}=\frac{2}{\sqrt 3}. $$

Answer 2

Just remove the lower order terms since their effect is negligible at infinity:

$$ \lim_{x\to\infty} \frac{2x^{1/2}+3x^{1/3}+5x^{1/5}}{(3x-2)^{1/2}+(2x-3)^{1/3}} = \lim_{x\to\infty} \frac{2x^{1/2}}{(3x)^{1/2}} = \frac{2}{\sqrt{3}} $$