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How to find the limit of $ \frac{\tan^2x - x^2}{x^2\tan^2x}$ as $x$ approaches $0$.

I don't know how to approach and starting differentiating the numerator and denominator many times which after the 4th time gave me the result $\frac{2}{3}$. I want to know if there is any good simplification which can lead me to a simple and quick solution. Please give some ideas.

Thank you.

Just_A_User
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4 Answers4

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Note that as $x \to 0$, we have $\tan(x) = x + \dfrac{x^3}{3} + O(x^5)$ (this is really a 2 line calculation). So, \begin{align} \tan^2(x) &= x^2 + \dfrac{2x^4}{3} + O(x^6) \end{align} Hence, \begin{align} \dfrac{\tan^2x - x^2}{x^2 \tan^2x} &= \dfrac{\frac{2x^4}{3} + O(x^6)}{x^4 + O(x^6)} \\ &= \dfrac{\frac{2}{3} + O(x^2)}{1 + O(x^2)}, \end{align} where I divided top and bottom by $x^4$. Thus, if we take the limit $x \to 0$, the answer is $\dfrac{2}{3}$.

By the way, observe that it is no coincidence you had to differentiate $4$ times, because as you can see in the above derivation, the Taylor polynomials of the numerator and denominator begin at the $x^4$ term, so it is precisely the ratio of these coefficients which you're calculating if you use something like L'Hopital's rule.

peek-a-boo
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In denominator, you can multiply and divide by $x^2$, that would eliminate your $\tan x$ in denominator as $\lim_{x\to 0}\frac{\tan x}{x}=1$.

In numerator, you may use series expansion of $\tan x=x+\frac{x^3}{3}$. You need not write next terms as the denominator has degree $4$. So, more powers of $x$ in numerator would make it zero.

With this approach also, you would get answer as $\frac23$.

Detailed steps:$$\lim_{x\to 0}\frac{(x+\frac{x^3}{3})^2-x^2}{x^4}=\lim_{x\to 0}\frac{x^2+\frac{2x^4}{3}-x^2}{x^4}=\frac23$$ (once again, powers of $x$ higher than $4$ are ignored.)

aarbee
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  • To be rigourous, this amounts to using Taylor-Young's formula at order $4$ for the tangent: $;\tan x =x+\frac{x^3}3+o(x^4)$. – Bernard May 22 '20 at 08:18
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$$L = \lim_{x \to 0} \dfrac{\tan^2 x - x^2}{x^4} = \lim_{x \to 0} \dfrac{\tan x - x}{x^3} \dfrac{x^2}{\tan^2 x} \left(\dfrac{\tan x}{x} + 1\right)$$

$\lim_\limits{x \to 0}\dfrac{\tan x - x}{x^3} = \dfrac{1}{3}$ as evaluated here, we get $L = \dfrac{2}{3}$.

user8277998
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Recall that $\lim_{x\to 0}\dfrac{\sin(x)}{x}=1$. The proof of this involves the definition of trigonometric functions by unit circle, which I will omit here. An immediate consequence of this is $\lim_{x\to 0}\dfrac{\tan(x)}{x}=1$. Hence dividing both the numerator and the denominator by $x^2$, we see that the desired limit is indeed 0.

Divide1918
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