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Below i am trying to differentiate $\sin x$ using first principle.

$$\lim_{h \to 0} \frac{\sin(x+h)-\sin x}{h}$$ $$=\lim_{h \to 0} \frac{\sin x(\cos h-1)} {h}+\cos x\lim_{h \to 0} \frac{\sin h}{h}$$

This is a bit uncoventional from the way it is usually done in textbooks using the $\sin C-\sin D$ formula. My concern is the first term in the third line,which is

$$\lim_{h \to 0} \frac{\sin x(\cos h-1)}{h}=-\sin x \lim_{h \to 0} \frac{2\sin^2{\frac{h}{2}}}{h}$$

When evaluating this limit,is it okay to split the limit $$\lim_{h \to 0} \frac{2\sin^2{\frac{h}{2}}}{\frac{h^2}{2}} \times \lim_{h \to 0} \frac{h}{2}=2\times 0=0$$ as shown?

In other words,is it always okay to split limits as $$\lim_{x \to a} f(x) \times g(x)=\lim_{x \to a} f(x) \times \lim_{x \to a} g(x)=L_{f(x)}\times L_{g(x)}$$

as long as both $L_{f(x)}$ and $L_{g(x)}$ don't turn out to be undefined EVEN if both or either of $L_{f(x)}, L_{g(x)}$ are/is $0$? Please enlighten me.

madness
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2 Answers2

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Yes, it is correct to split the limit of the product into a product of limits, as long as both limits are well-defined and as long as you are not in the case $0\times\pm\infty$. All other cases (finite$\times$finite, $\pm\infty\times\pm\infty$, non-zero$\times\pm\infty$) are fine.

See here for a previous answer on this topic.

Andreas Tsevas
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Perhaps an alternative way is by using differentials.

From trigonometry: $$\sin(x + dx)= \sin(x)\cos(dx) + \cos(x)\sin(dx)$$

Since $\cos(dx)\approxeq 1$ and $\sin(dx)\approxeq dx$, we have:

$$\sin(x + dx)= \sin(x) + \cos(x)dx$$

But we know that the derivative of a function is the linear term in the infinitesimal $dx$ when expressing the function with Taylor series.

So $f^\prime(x)= \cos(x)$

yes
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