$$\lim\limits_{x \to \infty}\frac{\sqrt{64x^{12}+45x^5-100}}{6x^6-x^4+14}$$ What I mean by algebraically, is by doing transformations or other things, not just evaluating the function with excessively big numbers.
Also, I know that the limit is $\displaystyle\frac{4}{3}$. I just want to know how to obtain it. Thanks.
The standard method when tackeling such limits, is to factor out the highest powers of $x$.
You have
$\sqrt{64x^{12}+45x^5-100}=\sqrt{x^{12}\left(64+\dfrac{45}{x^7}-\dfrac{100}{x^{12}}\right)}=x^6\sqrt{64+\dfrac{45}{x^7}-\dfrac{100}{x^{12}}}$
and for the denominator
$x^6\left(6-\dfrac{1}{x^2}+\dfrac{14}{x^6}\right)$.
Now you cancel these powers. In this case it is kind off a coincidence, that the power of $x$ equals. This is also the most interesting case.
The idea behind this "method" is that to use the limit theorems, and now we can see the limit in the numerator and denominator easily, as the summands are either constant, or null sequences.
So the limit in the numerator is $\sqrt{64}$.
The limit in the denominator is $6$.
Then the limit of the fraction is $8/6=4/3$.