$lim x \to 0 \frac{{tan}^{2} [x]}{[x]^{2}}$ , where $[]$ represents greatest integer function, is?

1

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- 1

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0

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Does not exist

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Solution

The correct option is D $1$
First take the Right Hand Limit,

R.H.L.=

$R . H . L . = l i m h \to 0 \frac{{t a n}^{2} [h]}{{[h]}^{2}} = l i m h \to 0 \frac{{t a n}^{2} [0]}{{[0]}^{2}} = \frac{0}{0}$

which is meaningless

Now take, Left Hand Limit

$L . H . L . = l i m h \to 0 \frac{{t a n}^{2} [- h]}{{[- h]}^{2}} = l i m h \to 0 \frac{{t a n}^{2} [- 1]}{{[- 1]}^{2}} = \frac{{[t a n (1)]}^{2}}{1}$

Substitue value of $t a n (1)$

Thus

$L H L = {(1.557)}^{2} = 2.424$

As RHL does not exist and LHL is not equal to RHL

The given limit does not exist.

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