Proof that $\lim_{x\to0}\frac{\sin x}x=1$

Question

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.

Answer 1

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!