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We just completed limits in school, and I have quite a few doubts.

Limit x tends to 0 we found sin x=x and cos x=1-x. Subsequently we could substitute x as 0 in cos x to give 1 but not sin x to give 0. Can someone explain why?

Secondly, can someone explain why substituing a polynomial partially is wrong? Consider 2x/x as limit x tends to 0. We could write this is (x+x)/x and then substitue x=0 for the first x on the top to get x/x.

Thirdly, how can a fraction which initially starts in some form 0/0 or something like that evaluate to a finite number? Geometrically I understand it, but not algebraicaly.

Lastly, Can someone refer some sort of course/book which deals in limits in details and rigour from scratch?

jedi nerd
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1 Answers1

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Note that $\displaystyle \lim_{x \to 0} \dfrac{\sin x}{x}$ makes sense, and so does $\displaystyle \lim_{x \to 0} \dfrac{\cos x}{1-x}$, and both are equal to $1$. What this means is, when $x$ gets closer and closer to $0$, the expressions approach the value $1$. You can read the proof for the first limit here.

But when you write $\sin x=x$, this means you want a solution to this equation, that is, a value of $x$ which will satisfy this equation. I assume that you have read the indeterminate forms wikipedia page which I mentioned in comments. So, if an expression on substituting $x=0$, becomes of the form $0/0$ or any other indeterminate form mentioned, you cannot evaluate it's value at that point, simply because it's not defined at that point. What then you can do is, try and see what value does the expression approaches when $x$ approaches $0$, because it is defined in it's neighbourhood, and that is what it means to take the limit at that point.

You cannot substitute $x=0$ partially in an expression. It's totally wrong. You have to replace every $x$ by a $0$ while substituting.

A good book for starting limits would be "Essential Calculus: Early Transcendentals by James Stewart".

V.G
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