I need to find the limit of $$\lim_{x\to \infty}\left(x-\sqrt{\frac{4x^3+3x^2}{4x-3}}\right)$$ I did rooting like this: $$\frac{\left(x-\sqrt{\frac{4x^3+3x^2}{4x-3}}\right)*\left(x+\sqrt{\frac{4x^3+3x^2}{4x-3}}\right)}{x+\sqrt{\frac{4x^3+3x^2}{4x-3}}} = \frac{x^2-\frac{4x^3+3x^2}{4x-3}}{x+\sqrt{\frac{4x^3+3x^2}{4x-3}}}$$ but I got stuck. How can I continue from here? Thanks in advance.
2026-04-08 04:31:27.1775622687
How do I solve this limit $\lim_{x\to \infty}\left(x-\sqrt{\frac{4x^3+3x^2}{4x-3}}\right)$?
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Putting the terms in the numerator over the common denominator one obtains: $$\begin{align}\frac{x^2-\frac{4x^3+3x^2}{4x-3}}{x+\sqrt{\frac{4x^3+3x^2}{4x-3}}}&=\frac{-6x^2}{(4x-3)x\left(1+\sqrt{\frac{4x+3}{4x-3}}\right)}\\ &=\frac{-6}{(4-\frac3{x})\left(1+\sqrt{\frac{4x+3}{4x-3}}\right)}. \end{align}$$
Can you get it from here?