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Our calculus book states elementary limits like $\lim_{x\to0} \frac{\sin x}{x}=1$, or $\lim_{x\to0} \frac{\ln (1+x)}{x}=1$, $\lim_{x\to0} \frac{e^x-1}{x}=1$ without proof.

At the end of the chapter of limits, it shows that these limits can be evaluated by using series expansion (which is not in our high school calculus course).

However, series expansion of a function can only be evaluated by repeatedly differentiating it.

And, to calculate derivative of $\sin x$, one must use the $\lim_{x\to0} \frac{\sin x}{x}=1$.

So this seems to end up in a circular logic. It is also same for such other limits.

I found that $\lim_{x\to0} \frac{e^x-1}{x}=1$ can be proved using binomial theorem.

How to evaluate other elementary limits without series expansion or L'Hôpital Rule?

This answer does not explain how that limit can be evaluated.

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  • $\begingroup$ Can you give us a proof of the assertion "..series expansion of a function can only be evaluated by repeatedly differentiating it."? $\endgroup$
    – user170039
    Commented May 12, 2017 at 5:26
  • $\begingroup$ I wanted to mean Taylor series or Maclaurin Series, not binomial expansion. Should I update the question that way? One can write the power series $$ f(x) = a_0 + a_1(x-a) + a_2(x-a)^2 + \dots + a_n(x-a)^n + \dots $$ , and differentiate both sides to evaluate $a_i$ $\endgroup$
    – Archisman Panigrahi
    Commented May 12, 2017 at 5:27
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    $\begingroup$ You can say that these are circular if the functions are defined via some means other than series. If these are defined via series then there is no circularity. Moreover there are many choices to define these functions and each has its own advantages and challenges. None of these is more correct than the other. So it all boils down to how you define them. $\endgroup$
    – Paramanand Singh ♦
    Commented May 12, 2017 at 6:32
  • $\begingroup$ Moreover note that the proper development of the theory of these functions (exponential, logarithmic and circular) is a non trivial task and mostly not suitable for a beginner in calculus. It is preferable to be intellectually honest and mention the properties of these functions including these limits without proof and say that proofs will be provided in advanced courses. On the hand if a textbook on introductory calculus proves these limits via L'Hospital's Rule or Taylor's series then that is an intellectual fraud. $\endgroup$
    – Paramanand Singh ♦
    Commented May 12, 2017 at 6:49
  • $\begingroup$ Do have a look at the blog posts paramanands.blogspot.com/2014/05/… and paramanands.blogspot.com/2016/03/… if you are interested in the theory of these functions. $\endgroup$
    – Paramanand Singh ♦
    Commented May 12, 2017 at 6:53

1 Answer 1

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Hint to prove it yourself:

enter image description here

Let $A_1$ be the area of triangle $ABC$, $A_2$ be the area of arc $ABC$, and $A_3$ be the area of ABD. Then we have:

$$A_1<A_2<A_3$$

Try and find expressions for $A_1,A_2,$ and $A_3$ and fill them into the inequality and finally use the squeeze theorem.

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  • $\begingroup$ The figure here seems to be same as that in the classic answer by moderator rob John. Do visit math.stackexchange.com/a/75151/72031 for more discussion on this limit. $\endgroup$
    – Paramanand Singh ♦
    Commented May 12, 2017 at 6:40
  • $\begingroup$ Just realized this. I googled image illustration for sinx/x proof and this one was the best, didn't know it was from math stack exchange. $\endgroup$
    – AspiringMat
    Commented May 12, 2017 at 6:47
  • $\begingroup$ The advantage of this approach is that it can be presented with least amount of mathematical machinery and is suitable for students who are beginning to learn calculus. The only deep result needed is that every sector of a circle has an area and that area of a region is not less than area of a sub-region. Also contrary to what many believe the proof is as rigorous as a proof can be. $\endgroup$
    – Paramanand Singh ♦
    Commented May 12, 2017 at 6:57

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