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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. $\large \lim\limits_{x\to0}\dfrac{x}{\tan(7x)}$
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  • $\begingroup$ do you know the value of $\tan(0)$ ? $\endgroup$ – reuns Jul 21 '15 at 23:10
  • $\begingroup$ @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. $\endgroup$ – Sam Perales Jul 21 '15 at 23:13
  • $\begingroup$ do you know the value of $\tan'(0)$ ? do you know a theorem about $\lim_{x\to 0} \frac{f'(x)}{x}$ ? $\endgroup$ – reuns Jul 21 '15 at 23:19
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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 $$ \lim_{\large x \to 0}\frac{\tan x}x=1. $$

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  • $\begingroup$ @Oliver , Why are you multiplying by 1/7? $\endgroup$ – Sam Perales Jul 21 '15 at 23:17
  • $\begingroup$ @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. $\endgroup$ – Olivier Oloa Jul 21 '15 at 23:20
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    $\begingroup$ Oh, I see. The main goal was to get the tan x to the top. $\endgroup$ – Sam Perales Jul 21 '15 at 23:22
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The difference is if you have to worry about asymptotes. the first one the limit evaluated as x approaches 0 from the right. It doesn't worry about the left portion of the limit. However, the second one includes what happens as x approaches 0 from the left side as well as the right.

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