The value of \(\rm \displaystyle\lim_{x\to 0}\dfrac{\tan x - x}{x^2 \tan x}\) is equal to:

Concept:​

• \(\rm \displaystyle \lim_{x\to0}\dfrac{\tan x}{x}=1\).
• \(\rm \dfrac{d}{dx}\tan x=\sec^2x\).
• \(\rm \dfrac{d}{dx}\sec x=\tan x\sec x\).
• \(\rm \dfrac{d}{dx}\left[f(x)\times g(x)\right]=f(x)\dfrac{d}{dx}g(x)+g(x)\dfrac{d}{dx}f(x)\).

Indeterminate Forms: Any expression whose value cannot be defined, like \(\dfrac00\), \(\pm\dfrac{\infty}{\infty}\), 0⁰, ∞⁰ etc.

• For the indeterminate form \(\dfrac 0 0\), first try to rationalize by multiplying with the conjugate, or simplify by cancelling some terms in the numerator and denominator. Else, use the L'Hospital's rule.
• L'Hospital's Rule: For the differentiable functions f(x) and g(x), the \(\rm \displaystyle \lim_{x\to c} \dfrac{f(x)}{g(x)}\), if f(x) and g(x) are both 0 or ±∞ (i.e. an Indeterminate Form) is equal to the \(\rm \displaystyle \lim_{x\to c} \dfrac{f'(x)}{g'(x)}\) if it exists.

Calculation:

\(\rm \displaystyle\lim_{x\to 0}\dfrac{\tan x - x}{x^2 \tan x}=\dfrac00\) is an indeterminate form. Let us simplify and use the L'Hospital's Rule.

\(\rm \displaystyle\lim_{x\to 0}\dfrac{\tan x - x}{x^2 \tan x}=\lim_{x\to 0}\left[\dfrac{\tan x - x}{x^3}\times\dfrac{x}{\tan x}\right]\).

We know that \(\rm \displaystyle\lim_{x\to 0}\dfrac{x}{\tan x}=1\), but \(\rm \displaystyle \lim_{x\to 0}\dfrac{\tan x - x}{x^3}\) is still an indeterminate form, so we use L'Hospital's Rule:

\(\rm \displaystyle \rm \displaystyle \lim_{x\to 0}\dfrac{\tan x - x}{x^3}=\lim_{x\to 0}\dfrac{\sec^2 x - 1}{3x^2}\), which is still an indeterminate form, so we use L'Hospital's Rule again:

\(\rm \displaystyle \lim_{x\to 0}\dfrac{\sec^2 x - 1}{3x^2}= \lim_{x\to 0}\dfrac{2\sec x(\sec x\tan x)}{6x}=\lim_{x\to 0}\dfrac{\sec^2 x\tan x}{3x}\), which is still an indeterminate form, so we use L'Hospital's Rule again:

\(\rm \displaystyle \lim_{x\to 0}\dfrac{\sec^2 x\tan x}{3x}=\lim_{x\to 0}\dfrac{\sec^2 x\sec^2 x+\tan x[2\sec x(\sec x \tan x)]}{3}=\dfrac{1}{3}\).

∴ \(\rm \displaystyle\lim_{x\to 0}\dfrac{\tan x - x}{x^2 \tan x}=1\times\dfrac{1}{3}=\dfrac{1}{3}\).