1
$\begingroup$

$$\lim_{x \to 0} \frac{\tan^{12}x-x^{12}}{x^{14}} $$

I tried writing it as $\lim\limits_{x \to 0} \dfrac{\dfrac{\tan^{12}x}{x^{12}}-1}{x^{2}} $ thinking that if I applied L'hospital two times I would reach my result but with no luck. Any suggestions or tricks I could use?

$\endgroup$
  • $\begingroup$ $\tan^{12}$ means $\tan^{12} x$? $\endgroup$ – астон вілла олоф мэллбэрг Jun 14 '18 at 3:38
  • $\begingroup$ yeah forgot to add x $\endgroup$ – The Virtuoso Jun 14 '18 at 3:39
  • $\begingroup$ Don't do L'Hopital 12 times. It doesn't help much anyway in this problem but $$\frac{\tan^{12}}{x^{12}}=\left( \frac{\tan x}x\right)^{12}$$ since the limit $L$ of the inside exists and $x^{12}$ is continuous, the limit of the whole fraction is $L^{12}$ $\endgroup$ – N8tron Jun 14 '18 at 3:48
  • $\begingroup$ I am writing an answer. I request you to wait for the same. $\endgroup$ – астон вілла олоф мэллбэрг Jun 14 '18 at 4:01
  • 1
    $\begingroup$ The result should be 4 $\endgroup$ – The Virtuoso Jun 14 '18 at 4:08
3
$\begingroup$

It is best to first establish the well known limit $$\lim_{x\to 0}\frac{\tan x-x} {x^3}=\frac{1}{3}\tag{1}$$ This is easily done via the application of L'Hospital's Rule or Taylor series expansions.

The exponent $12$ is used here to intimidate students and one just needs to replace it with a generic symbol $n$. Next we note that for any positive integer $n$ we have $$f(n) =\lim_{x\to 0}\frac{\tan^n x-x^n} {x^{n+2}}=\lim_{x\to 0}\frac{\tan x-x} {x^3}\cdot\sum_{i=1}^{n}\frac{\tan^{i-1}x}{x^{i-1}} =\frac{n}{3}$$ The limit in question is $f(12)=4$.

Thus if one wishes to use L'Hospital's Rule then just a single application of the rule is fine. But one must always use a certain amount of algebraic manipulation before applying the rule.


In case $n$ is not a positive integer then it is best to make use of another standard limit $$\lim_{x\to a} \frac{x^n-a^n} {x-a} =na^{n-1}\tag{2}$$ along with limit $(1)$. Thus we have $$f(n) =\lim_{x\to 0}\frac{\tan^n x-x^n} {x^{n+2}}=\lim_{x\to 0}\dfrac{\left(\dfrac{\tan x} {x} \right) ^n-1}{\dfrac{\tan x} {x} - 1}\cdot\dfrac{\dfrac{\tan x} {x} - 1}{x^2}\\=\lim_{t\to 1}\frac{t^n-1}{t-1}\cdot\lim_{x\to 0}\frac{\tan x-x} {x^3}=\frac{n}{3}$$ Here we have used the substitution $t=(\tan x) /x$ so that $t\to 1$ as $x\to 0$.

$\endgroup$
  • $\begingroup$ (+1) I was writing up an answer that was very close to your first argument. $\endgroup$ – robjohn Jun 14 '18 at 6:44
2
$\begingroup$

Hint

The simplest would be to use Taylor series $$\tan(x)=x+\frac{x^3}{3}+O\left(x^4\right)$$ $$ \frac{\tan^{12}(x)-x^{12}}{x^{14}}=\frac{x^{12}\left(1+\frac{x^2}{3}+O\left(x^3\right) \right)^{12} -x^{12}}{x^{14} }=\frac 1 {x^2}\left(\left(1+\frac{x^2}{3}+O\left(x^3\right) \right)^{12} -1\right)$$ Now, use the binomial expansion.

Even simpler, remember that, for small $\epsilon$, $(1+\epsilon)^n \sim 1+n\epsilon$

$\endgroup$
  • 2
    $\begingroup$ Sorry but I don't know how to use Taylor series I need an answer for high school level (if there exists one) $\endgroup$ – The Virtuoso Jun 14 '18 at 3:50
  • $\begingroup$ @TheVirtuoso Impress your teacher or prof by using the Taylor series man! Guessing you are a junior or senior...your teacher will likely be very impressed. $\endgroup$ – Prime Jun 14 '18 at 3:53
  • $\begingroup$ Junior or senior should know the important power series such as for log, exp, sin, cos, $(1+x)^n, (1-x)^{-n}$. $\endgroup$ – marty cohen Jun 14 '18 at 4:25
1
$\begingroup$

$\lim_{x \to 0} \frac{\tan^{12}x-x^{12}}{x^{14}} $

$\begin{array}\\ \lim_{x \to 0} \dfrac{\tan^{12}x-x^{12}}{x^{14}} &=\lim_{x \to 0} \dfrac{(\tan(x)/x)^{12}-1}{x^{2}}\\ &=\lim_{x \to 0} \dfrac{(\sin(x)/(x\cos(x))^{12}-1}{x^{2}}\\ &=\lim_{x \to 0} \dfrac{(\frac{\sin(x)}{x\cos(x)})^{12}-1}{x^{2}}\\ &=\lim_{x \to 0} \dfrac{(\frac{x-x^3/6+O(x^5)}{x(1-x^2/2+O(x^4))})^{12}-1}{x^{2}}\\ &=\lim_{x \to 0} \dfrac{(\frac{x-x^3/6+O(x^5)}{x-x^3/2+O(x^5))})^{12}-1}{x^{2}}\\ &=\lim_{x \to 0} \dfrac{(\frac{1-x^2/6+O(x^4)}{1-x^2/2+O(x^4)})^{12}-1}{x^{2}}\\ &=\lim_{x \to 0} \dfrac{(1+x^2/3+O(x^4))^{12}-1}{x^{2}}\\ &=\lim_{x \to 0} \dfrac{1+4x^2+O(x^4)-1}{x^{2}}\\ &=\lim_{x \to 0} 4+O(x^2)\\ &=4\\ \end{array} $

$\endgroup$
  • 1
    $\begingroup$ Is it there no other way besides Taylor? This problem is from an entrance college exam from 2 years ago so there must be one $\endgroup$ – The Virtuoso Jun 14 '18 at 4:10
  • $\begingroup$ The problem is that $\sin(x)/x \to 1$ does not have enough precision. Additional terms in sin and cos are needed. $\endgroup$ – marty cohen Jun 14 '18 at 4:22
  • 2
    $\begingroup$ The question is fundamentally "what is the $x^3$ coefficient of the Maclaurin series of $\tan x$". @TheVirtuoso $\endgroup$ – Lord Shark the Unknown Jun 14 '18 at 4:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.