Can $\lim_{x \to \infty} = \frac{x^2}{x-1}$ be solved without L'hopital I need to find limit for:
$$\lim_{x \to \infty} = \frac{x^2}{x-1}$$
I could solve this easily with L'hopital's rule. Taking derivative of both numerator and denumatorator gives:
$$\lim_{x \to \infty} = \frac{2x}{1} = \infty$$
But is there a way to solve this without using it? I tried multiplying both parts of the fraction with conjugate $(x+1)$, and then dividing with largest element. But I get zero in the denumerator.

If we multiply the numerator and denominator by the reciprocal of the highest power of $x$ in the denominator, we obtain
\begin{align*}
\lim_{x \to \infty} \frac{x^2}{x - 1} & = \lim_{x \to \infty} \frac{x^2}{x - 1} \cdot \frac{\frac{1}{x}}{\frac{1}{x}}\\
& = \lim_{x \to \infty} \frac{x}{1 - \frac{1}{x}}\\
& = \infty
\end{align*}
since the numerator grows without bound while the denominator approaches $1$.
Alternatively, observe that
$$\frac{x^2}{x - 1} = x + 1 + \frac{1}{x - 1} > x + 1$$
whenever $x > 1$.  Hence,
$$\lim_{x \to \infty} \frac{x^2}{x - 1} \geq \lim_{x \to \infty} (x + 1) = \infty$$
