How can I calculate this limit?
$$\lim_{x \to \frac{1}{4}} \frac{x^2}{\left(2\sqrt{x} - 1\right)^2}$$
I know that it is $+\infty$ as I saw answers from online calculators, but they always use exponential and logs or L'hopital's rule. Is there another way of solving it (factoring or rationalising)? Thanks in advance.
$$\begin{align} \lim_{x \to \frac{1}{4}} \frac{x^2}{\left(2\sqrt{x} - 1\right)^2}&\\&=\frac{(\frac{1}{4})^2}{\left(2\sqrt{\frac{1}{4}} - 1\right)^2}\\ &=\frac{\frac{1}{16}}{\left(2\cdot\frac{1}{2} - 1\right)^2}\\ &=\frac{\frac{1}{16}}{\left(1 - 1\right)^2}\\ &=\frac{\frac{1}{16}}{0^+}\\&=\frac{1}{0^+} \end{align}$$
This quantity will appoach $+\infty$.
See the graph