Evaluating $\lim_{x \rightarrow 0}\frac{3}{x}\bigg(\frac{1}{\tanh(x)}-\frac{1}{x}\bigg)$ using L'Hôpital's rule I need to evaluate the following limit (using L'Hôpital's rule).
$$\lim_{x \rightarrow 0}\frac{3}{x}\bigg(\frac{1}{\tanh(x)}-\frac{1}{x}\bigg)$$
I expressed $\tanh(x)$ as $\frac{\sinh(x)}{\cosh(x)}$:
$$\lim_{x \rightarrow 0}\frac{3}{x}\bigg(\frac{1}{\frac{\sinh(x)}{\cosh(x)}}-\frac{1}{x}\bigg)=\lim_{x \rightarrow 0}\frac{3}{x}\bigg(\frac{\cosh(x)}{\sinh(x)}-\frac{1}{x}\bigg)$$
$$=\lim_{x \rightarrow 0}\frac{3}{x} \bigg(\frac{x\cosh(x)-\sinh(x)}{x\sinh(x)}\bigg)$$
But I do not know how to continue.

EDIT:
Continuing,
$$\lim_{x \rightarrow 0}\frac{3}{x} \bigg(\frac{x\cosh(x)-\sinh(x)}{x\sinh(x)}\bigg)=\lim_{x \rightarrow 0}3 \bigg(\frac{x\cosh(x)-\sinh(x)}{x^2\sinh(x)}\bigg)$$
$$=3\lim_{x \rightarrow 0}\bigg(\frac{\cosh(x)+x\sinh(x)-\cosh(x)}{2x\sinh(x)+x^2\cosh(x)}\bigg)=3\lim_{x \rightarrow 0}\bigg(\frac{\sinh(x)}{2\sinh(x)+x\cosh(x)}\bigg)$$
$$=3\lim_{x \rightarrow 0}\bigg(\frac{\cosh(x)}{2\cosh(x)+\cosh(x)+x\sinh(x)}\bigg)=3\lim_{x \rightarrow 0}\bigg(\frac{\cosh(x)}{2\cosh(x)+\cosh(x)+x\sinh(x)}\bigg)$$
$$=3\lim_{x \rightarrow 0}\bigg(\frac{\cosh(x)}{3\cosh(x)+x\sinh(x)}\bigg)=3\times \frac{1}{3}=1$$

Your help would be appreciated. THANKS!
 A: Solution :
$$\lim _{x\to \:0}\left(\frac{3}{x}\left(\frac{1}{\tanh \left(x\right)}-\frac{1}{x}\right)\right)$$
Taking out constant :
$$\lim _{x\to a}\left[c\cdot f\left(x\right)\right]=c\cdot \lim _{x\to a}f\left(x\right)$$
We get :
$$=3\cdot \lim _{x\to \:0}\left(\frac{1}{x}\left(\frac{1}{\tanh \left(x\right)}-\frac{1}{x}\right)\right)$$
Simplify :
$$\frac{1}{x}\left(\frac{1}{\tanh \left(x\right)}-\frac{1}{x}\right)=\frac{x-\tanh \left(x\right)}{x^2\tanh \left(x\right)}$$
We get :
$$=3\cdot \lim _{x\to \:0}\left(\frac{x-\tanh \left(x\right)}{x^2\tanh \left(x\right)}\right)$$
Apply L'Hopital's Rule first time :
$$=3\cdot \lim _{x\to \:0}\left(\frac{1-sech ^2\left(x\right)}{2x\tanh \left(x\right)+sech ^2\left(x\right)x^2}\right)$$
Apply L'Hopital's Rule second time :
$$=3\cdot \lim _{x\to \:0}\left(\frac{2sech ^2\left(x\right)\tanh \left(x\right)}{-2x^2sech ^2\left(x\right)\tanh \left(x\right)+4xsech ^2\left(x\right)+2\tanh \left(x\right)}\right)$$
Apply L'Hopital's Rule third time :
$$=3\cdot \lim _{x\to \:0}\left(\frac{2\left(-2sech ^2\left(x\right)\tanh ^2\left(x\right)+sech ^4\left(x\right)\right)}{4x^2sech ^2\left(x\right)\tanh ^2\left(x\right)-2x^2sech ^4\left(x\right)-12xsech ^2\left(x\right)\tanh \left(x\right)+6sech ^2\left(x\right)}\right)$$
Plug the value of $x=0$ :
$$=3\cdot \frac{2\left(-2sech ^2\left(0\right)\tanh ^2\left(0\right)+sech ^4\left(0\right)\right)}{4\cdot \:0^2sech ^2\left(0\right)\tanh ^2\left(0\right)-2\cdot \:0^2sech ^4\left(0\right)-12\cdot \:0\cdot sech ^2\left(0\right)\tanh \left(0\right)+6sech ^2\left(0\right)}$$
Simplify the answer and we will get :
$$=1$$
So, we calculated that :
$$\lim _{x\to \:0}\left(\frac{3}{x}\left(\frac{1}{\tanh \left(x\right)}-\frac{1}{x}\right)\right)=1$$
