Possible Flaw in taking derivatives in a limit

Question

I was calculating some limits involving trigonometric functions, but I was stuck in the following limit for a while : $$\lim_{x\to 0} \frac {-\cos^2 x - \cos x + 2 }{x.\sin x} $$ I evaluated it as below: $$\begin{align} & \lim_{x\to 0} \frac {-\cos^2 x - \cos x + 2 }{x.\sin x}\\ &= \lim_{x\to 0} \frac {\frac {-\cos^2 x - cos x + 2}{x} }{\sin x}\\ &= \lim_{x\to 0} \frac {\frac {\left(-\cos^2 x - \cos x + 2\right) - \left(-\cos^2 0 - \cos 0 + 2\right)}{x - 0} }{\sin x}\\ &= \lim_{x\to 0} \frac {\frac {d}{dx}\left(-\cos^2 x - \cos x + 2\right) }{\sin x}\\ &= \lim_{x\to 0} \frac {2 \cos x . \sin x + \sin x }{\sin x}\\ &= \lim_{x\to 0} {2 \cos x + 1}\\ &= 3\end {align}$$ But the correct answer is actually$$\frac 32$$ so I want to know what is the flaw in my work, and is it possible in general to use this trick.. (my exemple may be just an exception).

Answer 1

In short: you cannot say $$ \lim_{x\to 0} \frac{\frac{f(x)-f(0)}{x-0}}{g(x)} = \lim_{x\to 0} \frac{f'(x)}{g(x)} $$ which is what you used in your third step. This is simply not true in general.

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To see this on a simpler example, take e.g. $f$ defined by $f(x)=x^2$ and $g(x)=x$.

Then $$ \lim_{x\to 0} \frac{\frac{f(x)-f(0)}{x-0}}{g(x)} = \lim_{x\to 0} \frac{\frac{x^2}{x}}{x} \lim_{x\to 0} 1 = 1 $$ while $$ \lim_{x\to 0} \frac{f'(x)}{g(x)} = \lim_{x\to 0} \frac{2x}{x} = 2. $$

Answer 2

You cannot just replace the tangent formula by the derivative ; doing so would mean that you are actually taking the limit when x -> 0, but in that case, you must also divide by $sin(0)$, which you cannot do obviously.

What you should do here is use equivalent functions, namely the following :

$sin(x) \underset{x \rightarrow 0}{=}x + o(x^2)$

$cos(x) \underset{x \rightarrow 0}{=}1 - \frac{x^2}{2} + o(x^3)$

Where $o(x^n)$ is a term that represents negligeable values. Technically speaking, $o(x^n)$ is a non-zero function $g$ so that $\frac{g(x)}{x^n} \underset{x \rightarrow 0}{\rightarrow} 0$. When replacing functions in products or divisions by their equivalents, you can ignore the $o(x^n)$ term, but not when replacing functions in sums or differences. For example, this is true :

$\frac{sin(x)cos(x)}{x} \underset{x \rightarrow 0}{=} \frac{x \times cos(x)}{x} = cos(x) \underset{x \rightarrow 0}{\rightarrow} 1$

But this is false :

$(1 + \frac{1}{n})^n \underset{n \rightarrow \infty}{=} (1 + 0)^n = 1$

In your case, just don't forget to include the negligeable terms when replacing terms of the sum with equivalent functions, and you should be good - remember than squaring is doing a product.

Answer 3

If the limits exist and the denominator is nonzero, then

$$ \lim_{x \to a} \frac{ \frac{f(x) - f(a)}{x-a} }{g(x) } = \frac{\lim_{x \to a} \frac{f(x) - f(a)}{x-a}}{\lim_{x \to a} g(x)} $$

and so you could conclude

$$ \lim_{x \to a} \frac{ \frac{f(x) - f(a)}{x-a} }{g(x) } = \frac{f'(a)} {\lim_{x \to a} g(x)}$$

But you are not in that situation, and that's not what you tried to conclude either.