In my text book it is written that if $$\lim_{x\to0}\;\frac{\cos(4x) + a\cos(2x) + b}{x^4}$$ is finite then $\frac{\cos(4x) + a\cos(2x) + b}{x^4}$ should be of the form $0/0$ and therefore $\cos4x + a\cos2x + b$ must be zero at $x=0$. I do not understand why. Please explain this to me.
$\endgroup$ 46 Answers
$\begingroup$Hint. One may use the Taylor series expansion, as $x \to 0$, $$ \cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}+O(x^6) $$ giving $$ \frac{\cos (4x)+a \cos(2x)+b}{x^4}=\frac{\color{red}{1+a+b}}{x^4}+\frac{\color{red}{-8-2a}}{x^2}+\frac{2 (16+a)}{3}+O(x^2) $$ then one may conclude with $$ \color{red}{1+a+b}=\color{red}{0}\quad \color{red}{-8-2a}=\color{red}{0}. $$
$\endgroup$ 4 $\begingroup$Suppose that you have 2 functions $f,g$ such that: $$\lim_{x\to 0} \frac{f(x)}{g(x)} = \lambda \in \mathbb R \text { and } \lim_{x\to 0} g(x) = 0.$$
Now, consider the function $h(x) = \frac{f(x)}{g(x)},$ so $\lim\limits_{x\to 0} h(x) = \lambda.$ Also, it is true that $f(x) = h(x) \cdot g(x).$ Hence: $$\lim_{x\to 0} f(x) = \lim_{x\to 0} [h(x) \cdot g(x)] =\lim_{x\to 0} h(x) \cdot \lim_{x\to 0} g(x) = \lambda \cdot 0 = 0. $$
Notice, in our case $f$ is a continuous function, thus, $\lim\limits_{x\to 0} f(x) = f(0) = 0 .$
$\endgroup$ 1 $\begingroup$Consider the general case $$\lim_{x\to a}\frac {f(x)}{g(x)}.$$
"Handwavingly", we can say that we must always have one of the following cases ($\pm$ not considered):
N.B. When I write "$\mathrm{finite}$", it is implied that I mean "$\mathrm{nonzero~finite}$".
- $f(x)\to \mathrm{finite}$ and $g(x)\to \mathrm{finite}\implies \frac{f(x)}{g(x)}\to \frac{\mathrm{finite}}{\mathrm{finite}}=\mathrm{finite}$
- $f(x)\to \infty$ and $g(x)\to \mathrm{finite}\implies \frac{f(x)}{g(x)}\to \frac{\infty}{\mathrm{finite}}=\infty$
- $f(x)\to \mathrm{finite}$ and $g(x)\to \infty\implies \frac{f(x)}{g(x)}\to \frac{\mathrm{finite}}{\infty}=0$
- $f(x)\to 0$ and $g(x)\to \mathrm{finite}\implies \frac{f(x)}{g(x)}\to \frac{0}{\mathrm{finite}}=0$
- $f(x)\to \mathrm{finite}$ and $g(x)\to 0\implies \frac{f(x)}{g(x)}\to \frac{\mathrm{finite}}{0}=\infty$
- $f(x)\to \infty$ and $g(x)\to 0\implies \frac{f(x)}{g(x)}\to \frac{\infty}{0}=\infty$
- $f(x)\to 0$ and $g(x)\to \infty\implies \frac{f(x)}{g(x)}\to \frac{0}{\infty}=0$
or
- $f(x)\to \infty$ and $g(x)\to \infty\implies \frac{f(x)}{g(x)}\to \frac{\infty}{\infty}=\mathbf{indeterminate}$
- $f(x)\to 0$ and $g(x)\to 0\implies \frac{f(x)}{g(x)}\to \frac{0}{0}=\mathbf{indeterminate}$.
In the OP's case, we know that the denominator is $0$ and the limit of the fraction is finite. So by simple elimination, the numerator should be $0$.
$\endgroup$ $\begingroup$What your book is trying to say that if limit of a fraction exists and limit of denominator is zero, then limit of numerator has to be zero. So for a fraction with denominator tending to zero, for it to have finite limit, numerator should also tend to zero. But not every for every finite limit.
$\endgroup$ $\begingroup$Since limit is finite and denominator is going to 0, so numerator should also go to 0 in order to apply L'Hop rule and we are guaranteed that limit is finite so it means rule applied here and so numerator has to go to 0 so to obtain $\frac{0}{0}$ form
$\endgroup$ $\begingroup$It's just that the denominator is obviously going to zero, and if the denominator of a fraction is going to zero, the only hope of having a finite limit is if the numerator also goes to zero. It's not that every function $\frac {f(x)}{g(x)}$ which has a limit at some point must have the form $\frac 0 0$, for example $\lim_{x\to 2} \frac x {x + 1}$ is finite but isn't of that form.
If however we already know that $g(x)\to 0$, then if $f(x)$ does anything other than converge to $0$ as well, the limit can't be finite.
- If $f(x)$ converges to infinity, then the limit is infinite.
- If $f(x)$ converges to a finite number, then the limit is infinite.
- If $f(x)$ does not converge, then the limit either doesn't exist or is infinite.
