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Is there any way to prove that $$\lim_{x\to0}\frac{\sin x}x=1$$ only multiplying both numerator and denominator by some expression?

I know how to find this limit using derivatives, L'Hopital's rule, Taylor series and inequalities. The reason I tried to find it using only multiplying both numerator and denominator and then canceling out indeterminate terms is because the most other limits can be solved using this method.

This is an example: $$\begin{align}\lim_{x\to1}\frac{\sqrt{x+3}-2}{x^2-1}=&\lim_{x\to1}\frac{x+3-4}{\left(x^2-1\right)\left(\sqrt{x+3}+2\right)}\\=&\lim_{x\to1}\frac{x-1}{(x+1)(x-1)\left(\sqrt{x+3}+2\right)}\\=&\lim_{x\to1}\frac{1}{(x+1)\left(\sqrt{x+3}+2\right)}\\=&\frac{1}{(1+1)\left(\sqrt{1+3}+2\right)}\\=&\frac18\end{align}$$ It is obvious that we firtly multiplied numerator and denominator by $\sqrt{x+3}+2$ and then canceled out $x-1$. So, in this example, we can avoid indeterminate form multiplying numerator and denominator by $$\frac{\sqrt{x+3}+2}{x-1}$$ My question is can we do the same thing with $\frac{\sin x}x$ at $x\to0$? I tried many times, but I failed every time. I searched on the internet for something like this, but the only thing I found is geometrical approach and proof using inequalities and derivatives.

Edit
I have read this question before asking my own. The reason is because in contrast of that question, I do not want to prove the limit using geometrical way or inequalities.

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    You can use L'Hopital's Rule or use a Mclaurin Series. Trigonometric functions are not algebraic, so no special cancelling can be done, at least in the same matter as you'd like. – David P Oct 08 '15 at 22:41
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    Possible duplicate of How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$? – Gaurav Ramanan Oct 08 '15 at 22:49
  • @GauravRamanan. I have read that question before asking this. In that question, OP said "Best would be a geometrical solution", but I mentioned in my question that I have already proved it in geometrical way. My question is different way of proving it. –  Oct 08 '15 at 22:53
  • @Mathematician171 Oops sorry, missed it! – Gaurav Ramanan Oct 08 '15 at 22:56
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    Using L'Hopital's rule or a McLaurin series isn't kosher, because those require knowing the derivative of sine, and proving that the derivative of sine is cosine is generally done using this limit. – Paul Sinclair Oct 08 '15 at 23:07
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    "I know how to find this limit using derivatives, L'Hopital's rule, Taylor series" That's a bit suspect. One usually proves $\lim \sin x /x = 1$ first, with a bare-handed geometry argument. From that you obtain the derivatives of the trig functions. Finding $\lim \sin x /x = 1$ using L'Hopital is a bit of joke really. – zhw. Oct 08 '15 at 23:10
  • "I searched on the internet for something like this, but the only thing I found is geometrical approach and proof using inequalities and derivatives." I doubt very much you found that. Inequalities yes, derivatives no. – zhw. Oct 08 '15 at 23:14
  • The problem with what you ask is the definition of $\sin$. The geometric "definition" doesn't give you a lot of wiggle room. If you use another definition, than that's a different matter. – Git Gud Oct 08 '15 at 23:42
  • "I do not want to prove the limit using geometrical way" So you want to prove a theorem about a geometry-defined function with no geometry. Unfortunately, I don't think there's any non-circular proof of the limit besides the famous geometry-based one. What don't you like about that proof? – J.G. Jan 24 '19 at 11:20

1 Answers1

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If I remember correctly, it is not possible to solve this limit algebraically, as therefore the Numerator $f(x)=\sin{x}$ and the Denominator $g(x)=\frac{1}x$ must have both an existing limit, so one can use the rule $$\lim_{x\to x_0}\frac{f(x)}{g(x)}=\frac{c}{d}$$

The right and lefthand side limits of the Denominator $g(x)=\frac{1}x$ are $$\lim_{x\to 0+}\frac{1}x=\infty$$ and $$\lim_{x\to 0-}\frac{1}x=-\infty$$

So you cannot solve this limit algebraically, as the limit is not defined at $x=0$.

Hope this helps!

mehtab_ahm
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