Did I solve this limit problem correctly? $\lim_{x \to 3}\left(\frac{6-x}{3}\right)^{\tan \frac{\pi x}{6}}$ Need to solve this limit
$$\lim_{x \to 3}\left(\frac{6-x}{3}\right)^{\tan \frac{\pi x}{6}}$$ 
When I put $x$ in this expression I have indeterminate form $1^{\infty }.$ 
 So I choose the way which was described on this page (2nd method)
http://www.vitutor.com/calculus/limits/one_infinity.html
$$e^{\lim_{x \to 3 }{\tan \frac{\pi x}{6}(\frac{6-x}{3}-1)}}=e^{\lim_{x \to 3 }{\tan \frac{\pi x}{6}(\frac{6-x}{3}-\frac{3}{3})}} = e^{\lim_{x \to 3 }{\tan \frac{\pi x}{6}(\frac{3-x}{3})}}$$
and now I just put $x$ in this expression and I got $$e^{\infty \cdot 0} = e^0 =1$$
P.S. If you don't see my formulas, then  

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 A: With your 2nd method the limit is
$$ A := e^{\lim_{x \to 3 }{\tan \frac{\pi x}{6}(\frac{3-x}{3})}}
=e^{\lim_{x \to 3 }\frac{\frac{3-x}{3}}{\cot \frac{\pi x}{6}}}
$$
Now use l'Hôpital's rule
$$
 A = e^{\lim_{x \to 3 }\frac{\frac{3-x}{3}}{\cot \frac{\pi x}{6}}}
= e^{\frac{-\frac{1}{3}}{-\frac{\pi}{6}}} = e^{\frac{2}{\pi}} 
$$
A: Your solution is correct halfway:
$$\lim_{x \to 3}\left(\frac{6-x}{3}\right)^{\tan \frac{\pi x}{6}}=\exp\left\{\lim_{x \to 3 }{\tan \frac{\pi x}{6}\left(\frac{6-x}{3}-1\right)}\right\}$$ Above you used two facts: 


*

*$\lim(\exp)=\exp(\lim)$, which is the continuity of exponent

*$\frac{1+t}{t}\to1$ if $t\to0$: $$\lim_{x\to3} \tan\frac{\pi x}{6}\ln \frac{6-x}{3}=$$ $$\lim_{x\to3} \tan\frac{\pi x}{6}\ln\left(1+\frac{6-x}{3}-1\right)=\lim_{x\to3} \tan\frac{\pi x}{6}\ln\left(1+\frac{6-x}{3}-1\right)=$$ $$\lim_{x\to3} \tan\frac{\pi x}{6}\left(\frac{6-x}{3}-1\right)\frac{\ln\left(1+\frac{6-x}{3}-1\right)}{\frac{6-x}{3}-1}=$$ $$\lim_{x\to3} \tan\frac{\pi x}{6}\left(\frac{6-x}{3}-1\right)\cdot \underbrace{\lim_{t\to0}\frac{\ln\left(1+t\right)}{t}}_1$$


Here we set $t=\frac{6-x}{3}-1\to0$ as $x\to 3$
Now it remains to calculate $$\lim_{x\to3} \tan\frac{\pi x}{6}\left(\frac{6-x}{3}-1\right)=\lim_{x\to3} \tan\frac{\pi x}{6}\frac{3-x}{3}=\frac{1}{3}\lim_{x\to3} (3-x)\tan\frac{\pi x}{6}$$
if we try to substitute $x=3$ we will obtain $0\cdot\infty$, which is one of the indeterminate forms and we cannot immediately conclude that it is $0$. We must carefully proceed with addressing this indeterminate forms. Further computation can take may paths. For example, we can use the L'Hospital's Rule $$\lim_{x\to3} (3-x)\tan\frac{\pi x}{6}=\lim_{x\to3}\frac{3-x}{\cot\frac{\pi x}{6}}=\left[\frac{\infty}{\infty}\right]=\lim_{x\to3}\frac{(3-x)'}{\left(\cot\frac{\pi x}{6}\right)'}=$$ $$\lim_{x\to3}\frac{-1}{-\frac{\pi}{6}\csc^2\frac{\pi x}{6}}=\frac{6}{\pi}$$
We do not forget the $1/3$ and the $exp$ in front:
$$\lim_{x \to 3}\left(\frac{6-x}{3}\right)^{\tan \frac{\pi x}{6}}=\exp\frac{1}{3}\frac{6}{\pi}=e^\frac{2}{\pi}$$
