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Help me find the limit as x approaches 0 from the left: $\quad\lim\limits_{x \to 0} \dfrac {\tan x}{1+\sin x}$

To apply l'Hopital's rule, I am not able to get this to: $0/0$

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    $\begingroup$ Both the numerator and denominator are continuous, and the denominator is non-zero at $x=0$, hence the limit is just the value of the function evaluated at $x=0$. $\endgroup$ – copper.hat Oct 20 '14 at 2:46
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    $\begingroup$ The key point to recognise here is: that as you don't have a limit of an indeterminate form, you cannot and do not need to apply l'Hopital's rule. Just find the limit. $\endgroup$ – Graham Kemp Oct 20 '14 at 2:50
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You do not need l'Hopital's rule. The expression evaluates to $\frac{0}{1+0} = 0$

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    $\begingroup$ More than "do not need": one cannot apply it. $\endgroup$ – Timbuc Oct 20 '14 at 3:06
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The limit is clear,

$$\lim_{x\rightarrow 0}\frac{\tan x}{1+\sin x}=\frac{\tan 0}{1+\sin 0}=\frac{0}{1+0}=0.$$

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L'Hȏpital's rule is only applicable when you have something of the form $\frac00$. However, you have: $$\lim_{x\to0}\frac{\tan x}{1+\sin x}=\frac{(0)}{1+(0)}=\frac01=0$$ so you don't have to use L'Hȏpital. The limit equals $0$.

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Since $\frac{\tan x}{1+\sin x}$ is a continuous at $0$, then we have $\frac{0}{1+0}=0$.

As we know,

If $f(x)$ is continuous at $x=x_0$, then $\lim_{x\to x_0}f(x)=f(x_0)$.

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