# How do you find the limit of (3x+1/x) - (1/sinx) as x approaches 0 using l'hospital's rule?

Nov 9, 2016

You can't evaluate the last expression using L'Hôpital's rule as it is not of an indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$

Nov 9, 2016

This is a bit of a non-answer...

#### Explanation:

$\left(3 x + \frac{1}{x}\right) - \left(\frac{1}{\sin} x\right) = \frac{3 {x}^{2} \sin x + \sin x - x}{x \sin x}$

Let:

$\left\{\begin{matrix}f \left(x\right) = 3 {x}^{2} \sin x + \sin x - x \\ g \left(x\right) = x \sin x\end{matrix}\right.$

Let:

$I = \left(- 1 , 1\right)$

Then:

• $f \left(x\right)$ and $g \left(x\right)$ are differentiable on $I$

• ${\lim}_{x \to 0} f \left(x\right) = {\lim}_{x \to 0} g \left(x\right) = 0$

• $g ' \left(x\right) = \sin x + x \cos x \ne 0$ when $x \in I \text{\} \left\{0\right\}$

• ${\lim}_{x \to 0} \frac{f ' \left(x\right)}{g ' \left(x\right)} = {\lim}_{x \to 0} \frac{6 x \sin x + 3 {x}^{2} \cos x + \cos x - 1}{\sin x + x \cos x}$

which is again in the form $\frac{0}{0}$

So it's not clear that ${\lim}_{x \to 0} \frac{f ' \left(x\right)}{g ' \left(x\right)}$ exists and the new form is more complicated than the original problem.

So L'Hôpital's rule does not help us much.