# What is the difference between these two limits, one with $\lim\limits_{x\to0^{+}}$, the other with $\lim\limits_{x\to 0}$?

I don't need an exact answer, I just need to know how these two limits would affect the answer and if there is a huge difference on how they are worked out, if they have a different step-by-step solution.

1. $\large \lim\limits_{x\to0^{+}}\dfrac{x}{\tan(7x)}$
2. $\large \lim\limits_{x\to0}\dfrac{x}{\tan(7x)}$
• do you know the value of $\tan(0)$ ? – reuns Jul 21 '15 at 23:10
• @reuns Yes, it is 0, I just need to know if I can still use the Limit Laws on #1 as I would on #2. – Sam Perales Jul 21 '15 at 23:13
• do you know the value of $\tan'(0)$ ? do you know a theorem about $\lim_{x\to 0} \frac{f'(x)}{x}$ ? – reuns Jul 21 '15 at 23:19

The function $$x \longmapsto f(x)=\frac{x}{\tan (7x)}$$ is even, thus in this case $$\lim_{\large x \to 0^-}\frac{x}{\tan (7x)} =\lim_{\large x \to 0^+}\frac{x}{\tan (7x)}=\lim_{\large x \to 0}\frac{x}{\tan (7x)}=\frac17\lim_{\large x \to 0}\frac{7x}{\tan (7x)}=\frac17$$ where we have used the standard result $$\lim_{\large x \to 0}\frac{\tan x}x=1.$$
• @SamPerales Because I know that $\lim_{\large u \to 0}\frac{\tan u}{u}=1$, and as I have $u=7x$, I need a factor $7$ to obtain $u$... Hoping it is clear now for you. Thanks. – Olivier Oloa Jul 21 '15 at 23:20
• Oh, I see. The main goal was to get the tan x to the top. – Sam Perales Jul 21 '15 at 23:22