Methods of Evaluating $\lim_{x\rightarrow 0} \frac{\sin x}{x}=1$ Multiple Choice Question Which of the following techniques for evaluating limits cannot be used to show $\lim_{x\rightarrow 0} \frac{\sin x}{x}=1$
$a:$ The Squeeze Theorem
$b:$ L'Hôpital's Rule
$c:$ Using the graph
$d:$ Factoring/Diving Out.
I want to ask is the answer $d$? I'm unsure whether it is $c$ or $d$.
 A: Option $a$ works because we can squeeze the limit using
$$ \cos(x)\le\frac{\sin(x)}{x}\le1.$$
I'll leave this to you, ask a question if it's unclear.
Option $b$ works if you are granted that $\frac{d\sin(x)}{dx}:=\cos(x)$. If not, then this is incorrect because it is circular reasoning: see Arthur's comment. If we are granted the definition of the derivative of $\sin(x),$ apply L'Hôpital's rule to the indeterminate form $\frac{0}{0}$. Differentiating top and bottom we get 
$$ \lim_{x\to0}\frac{\cos(x)}{1}=1.$$
Option $c$ works. Consider the graph: 
This indicates the limit as $x\to0$ is $1$.
Option $d$ however, does not work. We can't factor anything out of $\sin(x)$ such that a cancellation occurs. You can verify this.
A: I don't have enough rep. to comment yet. So, I will post this as an answer. I think you can use all methods. Using taylor expansion of sin(x) about zero, you can factor out an x from all terms, cancelling it with the one in the denominator, which will give the answer as 1 too. But I think your course has not yet covered taylor expansions. So, you could go with d. Please feel free to comment on my answer.
