4
$\begingroup$

I want to solve this limit:

$$\lim_{x\to-\infty}\sqrt{4x^2+3x}+2x$$

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

$$\lim_{x\to-\infty}\dfrac{3x}{\sqrt{4x^2+3x}-2x}$$

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

$\lim\limits_{x\to-\infty}\dfrac{3}{(4+\frac{3}{x})^{1/2}-2}$

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?

$\endgroup$
0
3
$\begingroup$

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 $-x\sqrt{4-3/x}$.

$\endgroup$
1
  • $\begingroup$ Of course!! I can't believe I forgot about something so simple! Thanks a bunch $\endgroup$
    – Xavier
    Oct 19 '15 at 15:26
3
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.