Question

Let [x] denote an integer less than or equal to x. Then:
$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\tan \left( \pi {{\sin }^{2}}x \right)+\left( |x|+{{\left( \sin \left( x|x| \right) \right)}^{2}} \right)}{{{x}^{2}}}$
(a) equals
(b) equals 0
(c) equals
(d) does not exist

Answer 1

We use the fact that we can evaluate limit $L$for real valued single variable function $f\left( x \right)$ at any point $x=a$ if and only if Left hand limit(LHL)= right hand limit(RHL)=the value of the function at $x=a$.We find left hand limit for given function $\displaystyle \lim_{x \to {{0}^{-}}}\dfrac{\tan \left( \pi {{\sin }^{2}}x \right)+{{\left( |x|-\sin x\left[ x \right] \right)}^{2}}}{{{x}^{2}}}$, right hand limit for given function $\displaystyle \lim_{x \to {{0}^{+}}}\dfrac{\tan \left( \pi {{\sin }^{2}}x \right)+{{\left( |x|-\sin x\left[ x \right] \right)}^{2}}}{{{x}^{2}}}$ and check whether they are equal.$$$$

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 $x=a$ 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 \displaystyle \lim_{x \to {{a}^{-}}}f\left( x \right)=\displaystyle \lim_{x \to {{a}^{+}}}f\left( x \right)=f\left( a \right) \\\ \end{aligned}$$ We are given the function to evaluate limit $$\displaystyle \lim_{x \to 0}\dfrac{\tan \left( \pi {{\sin }^{2}}x \right)+{{\left( |x|-\sin x\left[ x \right] \right)}^{2}}}{{{x}^{2}}}$$ Here $\left[ x \right] $ denotes the greatest integer function. We find the left hand limit first. $$\text{LHL=}\displaystyle \lim_{x \to {{0}^{-}}}\dfrac{\tan \left( \pi {{\sin }^{2}}x \right)+{{\left( |x|-\sin x\left[ x \right] \right)}^{2}}}{{{x}^{2}}}$$ We know that for $ x < 0 $the absolute value function is defined $\left| x \right|=-x $ and the sine function will lie in fourth quadrant will return $-\sin x $. We know that when we approaching from the left to evaluate limit, the distance $0-x=h $ is infinitesimally small positive quantity. So $x $will lie in the interval $\left[ -1,0 \right) $ and where the function $\left[ x \right] $ will return greatest integer smaller than $x $ that is $-1 $. So we have, $$\begin{aligned} & \text{LHL=}\displaystyle \lim_{x \to {{0}^{-}}}\dfrac{\tan \left( \pi {{\sin }^{2}}x \right)+{{\left( -x-\sin \left( x\left( -1 \right) \right) \right)}^{2}}}{{{x}^{2}}} \\\ & =\displaystyle \lim_{x \to {{0}^{-}}}\dfrac{\tan \left( \pi {{\sin }^{2}}x \right)+{{\left( -x+\sin x \right)}^{2}}}{{{x}^{2}}} \\\ \end{aligned}$$ We use law of addition of limits and have $$=\displaystyle \lim_{x \to {{0}^{-}}}\dfrac{\tan \left( \pi {{\sin }^{2}}x \right)}{{{x}^{2}}}+\displaystyle \lim_{x \to {{0}^{-}}}\dfrac{{{\left( -x+\sin x \right)}^{2}}}{{{x}^{2}}}$$ Let us divide and multiply $\pi {{\sin }^{2}}x $ in numerator and denominator in the first limit while expanding the second to have, $$\begin{aligned} & =\displaystyle \lim_{x \to {{0}^{-}}}\dfrac{\tan \left( \pi {{\sin }^{2}}x \right)\pi {{\sin }^{2}}x}{{{x}^{2}}\times \pi {{\sin }^{2}}x}+\displaystyle \lim_{x \to {{0}^{-}}}\dfrac{{{x}^{2}}+{{\sin }^{2}}x-2x\sin x}{{{x}^{2}}} \\\ & =\displaystyle \lim_{x \to {{0}^{-}}}\dfrac{\tan \left( \pi {{\sin }^{2}}x \right)}{\pi {{\sin }^{2}}x}\times \pi \times \dfrac{\sin x}{x}\times \dfrac{\sin x}{x}+\displaystyle \lim_{x \to {{0}^{-}}}\left( 1+\dfrac{\sin x}{x}\times \dfrac{\sin x}{x}-2\dfrac{\sin x}{x} \right) \\\ \end{aligned}$$ We use law the law addition and multiplication limits to have $$\begin{aligned} & =\displaystyle \lim_{x \to {{0}^{-}}}\dfrac{\tan \left( \pi {{\sin }^{2}}x \right)}{\pi {{\sin }^{2}}x}\times \pi \times \displaystyle \lim_{x \to {{0}^{-}}}\dfrac{\sin x}{x}\times \displaystyle \lim_{x \to {{0}^{-}}}\dfrac{\sin x}{x} \\\ & +\displaystyle \lim_{x \to {{0}^{-}}}1+\displaystyle \lim_{x \to {{0}^{-}}}\dfrac{\sin x}{x}\times \displaystyle \lim_{x \to {{0}^{-}}}\dfrac{\sin x}{x}-2\displaystyle \lim_{x \to {{0}^{-}}}\dfrac{\sin x}{x} \\\ \end{aligned}$$ We use the standard limits $\displaystyle \lim_{x \to 0}\dfrac{\tan x}{x}=1,\displaystyle \lim_{x \to \infty }\dfrac{\sin x}{x}=1,\displaystyle \lim_{x \to 0}c=c $ where $c $ is constant to have the left hand limit as , $$\begin{aligned} & =\left( 1\times \pi \times 1\times 1 \right)+\left( 1+1\times 1-2 \right)=\pi +0=\pi \\\ & \therefore \text{LHL=}\displaystyle \lim_{x \to {{0}^{-}}}\dfrac{\tan \left( \pi {{\sin }^{2}}x \right)+{{\left( |x|-\sin x\left[ x \right] \right)}^{2}}}{{{x}^{2}}}=\pi \\\ \end{aligned}$$ Similarly we find the right hand limit. We have, $$\text{RHL=}\displaystyle \lim_{x \to {{0}^{+}}}\dfrac{\tan \left( \pi {{\sin }^{2}}x \right)+{{\left( |x|-\sin x\left[ x \right] \right)}^{2}}}{{{x}^{2}}}$$ We know that for $ x > 0 $the absolute value function is defined $\left| x \right|=x $ and the sine function will lie in the first or second quadrant to return $\sin x $. We know that when we approach from the right to evaluate limits, $x $ will lie in the interval $\left( 0,1 \right] $ and where the function $\left[ x \right] $ will return a greatest integer smaller than $x $ that is 0. So we have, $$\begin{aligned} & \text{=}\displaystyle \lim_{x \to {{0}^{+}}}\dfrac{\tan \left( \pi {{\sin }^{2}}x \right)+{{\left( -x-\sin \left( x\left( 0 \right) \right) \right)}^{2}}}{{{x}^{2}}} \\\ & =\displaystyle \lim_{x \to {{0}^{+}}}\dfrac{\tan \left( \pi {{\sin }^{2}}x \right)+{{\left( -x \right)}^{2}}}{{{x}^{2}}} \\\ & =\displaystyle \lim_{x \to {{0}^{+}}}\dfrac{\tan \left( \pi {{\sin }^{2}}x \right)+{{x}^{2}}}{{{x}^{2}}} \\\ \end{aligned}$$ Let us use the law of addition of limits and proceed to have, $$=\displaystyle \lim_{x \to {{0}^{+}}}\dfrac{\tan \left( \pi {{\sin }^{2}}x \right)}{{{x}^{2}}}+\displaystyle \lim_{x \to {{0}^{+}}}\dfrac{{{x}^{2}}}{{{x}^{2}}}$$ We have already evaluated the first limit as $\pi $ while calculating the left hand limit. The limiting value will not change as the limiting function $\dfrac{\tan \left( \pi {{\sin }^{2}}x \right)}{{{x}^{2}}} $ is even. So we have the right hand limit, $$\begin{aligned} & =\pi +1 \\\ & \therefore \text{RHL=}\displaystyle \lim_{x \to 0}\dfrac{\tan \left( \pi {{\sin }^{2}}x \right)+{{\left( |x|-\sin x\left[ x \right] \right)}^{2}}}{{{x}^{2}}}=\pi +1 \\\ \end{aligned}$$ We neither compare left and right hand limits and find they are nor equal. Hence the limit does not exist at $x=0 $. **So, the correct answer is “Option D”.** **Note:** We note that the trick here is know the definitions of piecewise defined function like absolute value function otherwise known as modulus function $\left| x \right| $ and greatest integer function $\left[ x \right] $ for less and greater than 0. We cannot use L’hospital rule as $\left| x \right| $ and $\left[ x \right] $ are not differentiable within the neighbourhood of $x=0 $. The law of addition of limit is given for two functions $\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 law of multiplication of limits is given by $\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) $.