I want to solve this limit:


My try was to multiply by the conjugate, which gave me


But then factoring $x$ out of the denominator and cancelling with the $x$ at the numerator gives me


The problem is, when I evaluate this limit, I get $\frac{3}{0}$, but my book says the limit should be $\frac{3}{4}$.

Can anyone see where I made my mistake?


2 Answers 2


When factoring $x$ out you should check the sign of $x$. You might a mistake when treating $\sqrt{4x^2+3x}$. Our $x$ is negative so $\sqrt{4x^2+3x}$ is equal to $-x\sqrt{4-3/x}$.

  • $\begingroup$ Of course!! I can't believe I forgot about something so simple! Thanks a bunch $\endgroup$
    – Xavier
    Commented Oct 19, 2015 at 15:26

Note that $x$ is negative (as $x \to -\infty$). So, $$ \frac{\sqrt{4x^2 + 3x}}{x} = \frac{\sqrt{4x^2 + 3x}}{-\sqrt{x^2}} = -\sqrt{4 + \frac 3x} $$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .