Limits of Sin/Exp as x Approaches 0

[SomeRandomGuy]

$$\lim_{x\rightarrow\zero} \frac{\sin(x)}{\exp(x)}$$

 

[quasar987]

Why wouldn't that be simply 0 ?

 

[SomeRandomGuy]

Why wouldn't that be simply 0 ?

That's what I'm wondering. I have been trying to solve this limit for about 20 minutes now. BTW, that is the limit as x goes to 0... I don't know how to get that into the latext graphic :/

 

[quasar987]

As you know, in Latex, the \ "operator" is only used to input a special comand.. so you don't need one before writing 0 because zero is just a number. So it's simply

lim_{x \rightarrow 0}

As for the justification of the limit, we have that the limit of a quotient is the quotient of the limit if the limit exists and if the function at the denominator is never zero at least past a certain aribitrarily large x.

the limit of sinx is sin0 = 0 since sin x is continuous over all real.

the limit of exp(x) is e^0 =1 also because e^x is continuous on all real.

Therefor the limit of the quotient is 0/1 = 0.

 

[quasar987]

A cool proof for the limit of sinx makes use of the identity $0 \leq |sinx| \leq |x|$ $\forall x \in \mathbb{R}$ because you then have that

$$\lim_{x \rightarrow 0} 0 = 0$$

and

$$\lim_{x \rightarrow 0} |x| = 0$$.

such that, by the sandwich theorem,

$$\lim_{x \rightarrow 0} |sinx| = 0$$

And also using the "theorem" according to which

$$\lim_{x \rightarrow x_0} f(x) = 0 \Leftrightarrow \lim_{x \rightarrow x_0} |f(x)| = 0$$

(as you can very easily prove using the epsilon-delta definition of limit), we find the answer:

$$|sinx| \rightarrow 0 \Rightarrow sinx \rightarrow 0$$