Question

Solve the limit

$\lim \limits_{x \to 0} \dfrac{ {x}^{2} - \tan^2x }{ {x}^{4} }$
​

Answer 1

EXPLANATION.

$\sf \implies \displaystyle \lim_{x \to 0} \dfrac{x^{2} - tan^{2} x}{x^{4} }$

As we know that,

Put the value of x = 0 in the equation and check their indeterminant form, we get.

$\sf \implies \displaystyle \lim_{x \to 0} \dfrac{(0)^{2} - tan^{2} (0)}{(0)^{4} }$

$\sf \implies \displaystyle \lim_{x \to 0} \dfrac{0}{0}$

As we can see that,

It is in the form of 0/0 indeterminant.

As we know that,

Formula of :

$\sf \implies \displaystyle \lim_{x \to 0} \dfrac{tan(x)}{x} = 0$

$\sf \implies \displaystyle \lim_{x \to 0} \bigg[ \dfrac{(x - tanx)(x + tan x)}{x^{4} } \bigg]$

$\sf \implies \displaystyle \lim_{x \to 0} \bigg[ \dfrac{(x - tanx)(x + tan x)}{x.x^{3} } \bigg]$

$\sf \implies \displaystyle \lim_{x \to 0} \bigg[ \dfrac{(x + tan x)}{x^{} } \bigg] \times \dfrac{d}{dx} \displaystyle \lim_{x \to 0} \bigg[ \dfrac{(x - tanx)}{x^{3} } \bigg]$

$\sf \implies \displaystyle \lim_{x \to 0} \bigg[ 1 + \dfrac{tan(x)}{x} \bigg] \times \displaystyle \lim_{x \to 0} \bigg[ \dfrac{(1 - sec^{2} x)}{3x^{2} } \bigg]$

$\sf \implies \displaystyle \lim_{x \to 0} (1 + 1) \times \dfrac{(- tan^{2} x)}{3x^{2} }$

$\sf \implies \displaystyle 2 \times \lim_{x \to 0} \bigg( \dfrac{tan(x)}{x} \bigg)^{2} \times \dfrac{-1}{3}$

$\sf \implies \dfrac{-2}{3}$

$\sf \implies \displaystyle \lim_{x \to 0} \bigg[ \dfrac{x^{2} - tan^{2} x}{x^{4} } \bigg]= \dfrac{-2}{3}$

                                                                                                               

MORE INFORMATION.

Standard limits.

$(1) \sf \displaystyle \lim_{x \to 0} \frac{sin(x)}{x} = \lim_{x \to 0} \frac{x}{sin(x)} = 1 \ \ ; \lim_{x \to 0} sin(x) = 0$

$(2) \sf \displaystyle \lim_{x \to 0} cos(x) = \lim_{x \to 0} \bigg( \frac{1}{cos(x)} \bigg) = 1$

$(3) \sf \displaystyle \lim_{x \to 0} \frac{tan(x)}{x} = \lim_{x \to 0} \frac{x}{tan(x)} = 1 \ \ ; \lim_{x \to 0} tan(x) = 0$

Answer 2

Given :-

$\quad \leadsto \quad \lim \limits_{\sf x \to 0} \sf \dfrac{ {x}^{2} - \tan^2x }{ {x}^{4} }$

To Find :-

Value of the limit

Solution :-

If you substitute directly , x = 0 . So , the limit will lead to indeterminate form $\sf \dfrac{0}{0}$ .So , consider given again ;

${\quad \leadsto \quad \displaystyle \lim_{\sf x \to 0} \sf \dfrac{ {x}^{2} - \tan^2x }{ {x}^{4} } }$

${ : \implies \quad \sf \displaystyle \sf \lim_{x \to 0} \dfrac{(x)² - (\tan \: x )²}{x \times x³}}$

${ : \implies \quad \sf \displaystyle \sf \lim_{x \to 0} \dfrac{(x + \tan \: x ) ( x - \tan \: x )}{x \times x³}}$

${ : \implies \quad \sf \displaystyle \sf \lim_{x \to 0} \bigg\{ \dfrac{x + \tan x}{x} \times \dfrac{x - \tan x}{x³} \bigg\} }$

${ : \implies \quad \sf \displaystyle \sf \lim_{x \to 0} \bigg\{ \bigg( \dfrac{x}{x} + \dfrac{\tan x}{x} \bigg) \times \bigg( \dfrac{x- \tan x}{x³} \bigg) \bigg\} }$

${ : \implies \quad \sf \displaystyle \sf \lim_{x \to 0} \bigg( \dfrac{x}{x} + \dfrac{\tan x}{x} \bigg) \times \displaystyle \sf \lim_{x \to 0} \bigg( \dfrac{x- \tan x}{x³} \bigg) }$

Now , Use L'hopital rule in 2nd limit .

${ : \implies \quad \sf \displaystyle \sf \lim_{x \to 0} \bigg( 1 + \dfrac{\tan x}{x} \bigg) \times \displaystyle \sf \lim_{x \to 0} \bigg\{ \dfrac{(x-tan x)'}{(x³)'} \bigg\} }$

${ : \implies \quad \sf ( 1 + 1 ) \times \displaystyle \sf \lim_{x \to 0} \bigg( \dfrac{1 - \sec² x}{3x²} \bigg) }$

${ : \implies \quad \sf 2 \times \displaystyle \sf \lim_{x \to 0} \bigg\{ \dfrac{- ( \sec² x - 1 ) }{3x²} \bigg\} }$

${ : \implies \quad \sf 2 \times \displaystyle \sf \lim_{x \to 0} \bigg( \dfrac{- \tan² x }{3x²} \bigg) }$

${ : \implies \quad \sf 2 \times - \dfrac{1}{3} \times \displaystyle \sf \lim_{x \to 0} \bigg( \dfrac{ \tan² x }{x²} \bigg) }$

${ : \implies \quad \sf 2 \times - \dfrac{1}{3} \times \displaystyle \sf \lim_{x \to 0} {\bigg\{ \bigg( \dfrac{ \tan \: x }{x} \bigg) \bigg\}}^{2} }$

${ : \implies \quad \sf 2 \times - \dfrac{1}{3} \times \displaystyle \sf (1)²}$

${ : \implies \quad \bf - \dfrac{2}{3}}$

${\quad \qquad {\pmb {\bf { \orange { \therefore {\underline {\underline { \displaystyle \lim_{\bf x \to 0} \bf \dfrac{x² - \tan² (x)}{x⁴} = - \dfrac{2}{3} }}}}}}}}$

Used Concepts :-

• $\displaystyle \bf \lim_{\bf x \to 0} \bf \dfrac{\tan x}{x} = 1$

• $\bf ( a + b ) ( a - b ) = a² -b²$

• $\bf \sec² \phi = \tan² \phi + 1$

• If any limit is of the indeterminate form . Then L'hopital rule is applicable which can be applied by Differentiating both the numerator and denominator .

• $\bf \dfrac{d}{dx} ( \tan \: x ) = \sec² x$

• $\bf \dfrac{d}{dx} ( x^{n}) = n \cdot x^{(n-1)}$

• $\displaystyle \bf \lim_{x \to c} ( pq ) = \lim_{x \to c } p \times \lim_{x \to c } q$