Question

Evaluate the following limiting value: $\displaystyle \lim_{x \to a},\dfrac{{{\tan }^{2}}x-{{x}^{2}}}{{{x}^{2}}{{\tan }^{2}}x}$$$$$

Answer 1

Convert the given function using laws of limits to standard forms so that you can use known formulas to evaluate it.

Complete step by step answer:
We know that limiting value for any real valued single variable function $f\left( x \right)$ when the variable $x$ approaches to real number $a$ in the domain $f\left( x \right)$ is denoted by
$\displaystyle \lim_{x \to a}\,f\left( x \right)=L$
Here $L$ is called the limit of the function.
The limit $L$ exists for real valued single variable function $f\left( x \right)$ at any point $x=a$ then if and only if Left hand limit(LHL)= right hand limit(RHL)=the value of the function at $x=a$. In symbols,

& \text{LHL}=\text{RHL=}f\left( a \right) \\\ & \Rightarrow \underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right) \\\ \end{aligned}$$ We know from the laws of limits for the limit for two real valued functions $f\left( x \right)$ and $g\left( x \right)$ at point $x=a$ for both functions then by law of addition in limits $$\displaystyle \lim_{x \to a}\,f\left( x \right)+\displaystyle \lim_{x \to a}\,g\left( x \right)=\displaystyle \lim_{x \to a}\,\left( f\left( x \right)+g\left( x \right) \right)$$ and by law of multiplication in limits $$\displaystyle \lim_{x \to a}\,f\left( x \right)\cdot \displaystyle \lim_{x \to a}\,g\left( x \right)=\displaystyle \lim_{x \to a}\,\left( f\left( x \right)\cdot g\left( x \right) \right)$$ By law of division in limits $$\dfrac{\displaystyle \lim_{x \to a}\,f\left( x \right)}{\displaystyle \lim_{x \to a}\,g\left( x \right)}=\displaystyle \lim_{x \to a}\,\left( \dfrac{f\left( x \right)}{g\left( x \right)} \right)$$ If in the law of division has both the functions in numerator and denominator in the form $\dfrac{0}{0},\dfrac{\infty }{\infty }$ then we use L’hospital rule and differentiate the functions in numerator and denominator . $$\displaystyle \lim_{x \to a}\,\dfrac{f\left( x \right)}{g\left( x \right)}=\displaystyle \lim_{x \to a}\,\dfrac{{{f}^{'}}\left( x \right)}{{{g}^{'}}\left( x \right)}$$ The required limit is $\displaystyle \lim_{x \to a}\,\dfrac{{{\tan }^{2}}x-{{x}^{2}}}{{{x}^{2}}{{\tan }^{2}}x}$. We use the algebraic identity of $$({{a}^{2}}-{{b}^{2)}}$$ , law of multiplication and division in numerator and proceed to, $$\begin{aligned} & \displaystyle \lim_{x \to a}\,\dfrac{{{\tan }^{2}}x-{{x}^{2}}}{{{x}^{2}}{{\tan }^{2}}x} \\\ & =\displaystyle \lim_{x \to a}\,\dfrac{\left( \tan x-x \right)\left( \tan x+x \right)}{{{x}^{4}}\left( \dfrac{{{\tan }^{2}}x}{{{x}^{2}}} \right)} \\\ & =\dfrac{\displaystyle \lim_{x \to a}\,\dfrac{\left( \tan x-x \right)\left( \tan x+x \right)}{{{x}^{4}}}}{\displaystyle \lim_{x \to a}\,\left( \dfrac{\tan x}{x} \right)\displaystyle \lim_{x \to a}\,\left( \dfrac{\tan x}{x} \right)} \\\ \end{aligned}$$ We also know the standard formula $\displaystyle \lim_{x \to a}\,\dfrac{\tan x}{x}=1$ and the law of multiplication . $$\displaystyle \lim_{x \to a}\,\dfrac{\left( \tan x-x \right)\left( \tan x+x \right)}{{{x}^{4}}}=\displaystyle \lim_{x \to a}\,\dfrac{\left( \tan x-x \right)}{{{x}^{3}}}\displaystyle \lim_{x \to a}\,\dfrac{\left( \tan x+x \right)}{x}$$ Now we shall use L’Hopital rule for the form $\dfrac{0}{0}$ to differentiate numerator and denominator with respect to $x$ and also the formula $\displaystyle \lim_{x \to a}\,\sec x=1$ $$\begin{aligned} & \displaystyle \lim_{x \to a}\,\dfrac{\left( \tan x-x \right)}{{{x}^{3}}}\displaystyle \lim_{x \to a}\,\dfrac{\left( \tan x+x \right)}{x} \\\ & =\displaystyle \lim_{x \to a}\,\dfrac{{{\sec }^{2}}x-1}{3{{x}^{2}}}\displaystyle \lim_{x \to a}\,\dfrac{{{\sec }^{2}}x+1}{1} \\\ & =2\displaystyle \lim_{x \to a}\,\dfrac{{{\sec }^{2}}x-1}{3{{x}^{2}}} \\\ & =\dfrac{2}{3}\displaystyle \lim_{x \to a}\,\dfrac{{{\tan }^{2}}x}{{{x}^{2}}}\left( \because {{\sec }^{2}}x+{{\tan }^{2}}x=1 \right) \\\ & =\dfrac{2}{3}\left[ \displaystyle \lim_{x \to a}\,\dfrac{\tan x}{x}\cdot \displaystyle \lim_{x \to a}\,\dfrac{\tan x}{x} \right]=\dfrac{2}{3} \\\ \end{aligned}$$ **So the value of evaluated limit is $\dfrac{2}{3}$.** **Note:** The question tests your knowledge of identities in basic limits. Careful solving of simultaneous equations, substitution and use of formula will lead us to the correct result. The question can also be framed to find with sine instead of tan.