+118608 Hi 315, you probably already know that I am not great at limits but i have looked at yours and i have a question.
$$\displaystyle \lim_{x\rightarrow 0^+}\left(sin\left(e^{-\frac{1}{x^2}}\right)\right)\\\\\\
=\displaystyle \lim_{x\rightarrow 0^+}\left(sin\left(\frac{1}{e^{\frac{1}{x^2}}}\right)\right)\\\\\\$$
I agree with you that this equal 0 but I don't understand why you added the middle step of =lim x tends to 0 of sine x. I mean i know that also tends to 0 but why have you put them together?
Is there something that I don't know about??
+118608 Hi 315, you probably already know that I am not great at limits but i have looked at yours and i have a question.
$$\displaystyle \lim_{x\rightarrow 0^+}\left(sin\left(e^{-\frac{1}{x^2}}\right)\right)\\\\\\
=\displaystyle \lim_{x\rightarrow 0^+}\left(sin\left(\frac{1}{e^{\frac{1}{x^2}}}\right)\right)\\\\\\$$
I agree with you that this equal 0 but I don't understand why you added the middle step of =lim x tends to 0 of sine x. I mean i know that also tends to 0 but why have you put them together?
Is there something that I don't know about??
+1832 I did this step because when x tends to 0 , $$\frac{1}{e^}\frac{1}{x^2}$$ is tends to zero , so I replaced it by x
+118608 Okay fair enough 315.
I don't know why the model answer would have that step either.
Once again it is true it just seems very strange to me (that it is put there)
I assume it is suppose to make the final answer easier to get to but it doesn't work for me.
Maybe I am missing something 
I wonder if Alan could shed any light on this. ?? Could you Alan? 
