# Justification of Evaluating Limits by Substitution

It confuses me that when finding a limit, the same function which may yield a zero in its denominator can go on to yield a real limit via a different approach. Is it not strange that the same function can resolve to two different outcomes? For example the following function: $$\frac{dy}{d\theta}=\lim_{\Delta\theta\to0} \dfrac{sin(\theta+\Delta\theta)-sin(\theta)}{\Delta\theta}$$ I understand if the function does not exist at the limit value then getting a divide by zero makes sense. How come in this case various trig substitutions get around the issue?

• Evaluating a limit by simple substitution is only allowed when the function is continuous. I am not sure what "two different outcomes" you are referring to. If it exists, a limit can only be one value. – Qudit Feb 23 '17 at 5:54
• Since the derivative of the sine function is the cosine function, the limit of the above expression for $\theta=2\pi$ would be 1 when $\Delta\theta$ approaches $0$. I don't understand conceptually how using the trig identity for the sum of two angles gets around the divide by zero problem - since the function does not exist at that point. – James Izzard Feb 23 '17 at 6:04
• It's because the limit depends on values very close to but never actually equal to zero. – Qudit Feb 23 '17 at 6:06
• It is the same manipulation you would have for $$\lim_{\Delta\theta\to1} \dfrac{\Delta\theta^2-1}{\Delta\theta-1}$$. Do you understand his this works? – N74 Feb 23 '17 at 6:10