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### Without using L'Hopitals rule, how do you get $\lim_{x \to 0} \frac{1 - \cos 5x}{\sin 3x}$?

Note that $$\frac{1-\cos 5x}{\sin 3x} = \frac{1-\cos 5x}{5x} \cdot \frac{5x}{3x} \cdot \frac{3x}{\sin 3x}$$ Now appeal to known special limits & limit properties.
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### $\lim_{x \rightarrow 0} \frac{(x^2 - x^3) ((x + 1)^\sqrt{2} - 1)}{\log(1 + 4 x) (1 - \cos(2 x))}$

$$\frac{(x^2 - x^3) ((x + 1)^\sqrt{2} - 1)}{\log(1 + 4 x) (1 - \cos(2 x))}=\frac{(x+1)^\sqrt2-1}x\cdot\frac{x^2}{1-\cos2x}\cdot\frac{x-x^2}{\log(1+4x)}\;(**)$$ and just as Paramanand told you twice, ...
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### $\lim_{x \rightarrow 0} \frac{(x^2 - x^3) ((x + 1)^\sqrt{2} - 1)}{\log(1 + 4 x) (1 - \cos(2 x))}$

This is one using series expansions for elementary functions  \lim_{x \rightarrow 0} \frac{(x^2 - x^3) ((x + 1)^\sqrt{2} - 1)}{\log(1 + 4 x) (1 - \cos(2 x))}=\lim_{x \rightarrow 0} \frac{x^2 ((x + 1)...
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### How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?

Not at all appropriate for a cal one course (or any course), but for the perverse, egregious silliness of it: A circle is parametrized by $\gamma(t) = (\cos t, \sin t)$. Now, geometrically, the ...
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