# How can I solve this without using the L'Hôpital's Rule?

$$\large\lim_{x\to -\frac{3}{2}} \frac{2x^2+5x+3}{|2x+3|}$$

How can I compute this without using L'Hôpital's Rule?

• factorize the quadratic – Paul Oct 3 '14 at 13:32
• L'Hospital's Rule is unsuitable here anyway. – André Nicolas Oct 3 '14 at 13:34
• (also, in general L'Hospital is a pretty big hammer, and very often is overkill... it shouldn't be the first impulse to use it, at least in my opinion). – Clement C. Oct 3 '14 at 13:35

$$2x^2+5x+3=(2x+3)(x+1)$$

Now for real $y,\; |y|=\begin{cases} y &\mbox{if } y\ge0 \\ -y & \mbox{if } y<0\end{cases}$

Observe that the left & the right limits are different, hence the limit does not exist

• Pointless. Let the questioner have a go. – Paul Oct 3 '14 at 13:44
• @Paul, Not sure if I've understood your point? – lab bhattacharjee Oct 3 '14 at 14:12

Since $$\large\lim_{x\to -\frac{3}{2}^+} \frac{2x^2+5x+3}{|2x+3|}=\large\lim_{x\to -\frac{3}{2}^+} \frac{2x^2+5x+3}{2x+3}=\large\lim_{x\to -\frac{3}{2}^+} x+1=-\frac12.$$

and $$\large\lim_{x\to -\frac{3}{2}^-} \frac{2x^2+5x+3}{|2x+3|}=\large\lim_{x\to -\frac{3}{2}^-} - \frac{2x^2+5x+3}{2x+3}=\large\lim_{x\to -\frac{3}{2}^-} -( x+1)=\frac12.$$

We can conclude this limit doesn't exist!