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Let $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{[x]}^{2}}}{{{x}^{2}}}=l\And \underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{\left[ x \right]}^{2}}}{{{x}^{2}}}=m$, where [ ] denotes greatest integer, then:
A. l exists but m does not
B. m exists but l does not
C. l & m both exist
D. neither l nor m exists

Answer
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Hint: Find LHL and RHL of both the limits $l=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{[x]}^{2}}}{{{x}^{2}}}$and $m=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{\left[ x \right]}^{2}}}{{{x}^{2}}}$.

Complete step-by-step answer:
Use the definition of [x] property to solve the problem.
Don’t evaluate exact limits just relate LHL and RHL.
We have given that,
$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{[x]}^{2}}}{{{x}^{2}}}=l\ldots \ldots (1)$
and
$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{\left[ x \right]}^{2}}}{{{x}^{2}}}=m\ldots \ldots (2)$
where we need to find about l and m i.e. Will they exist or not?$\dfrac{0}{0}$
Now, as we know for the existence of any limit, LHL (Left hand limit) and RHL (Right hand limit) should be equal for any given limit expression.
Hence we need to find LHL and RHL for both the equations.
We know that [x] is the greatest integer for any given ‘x’ and take the left most integer value w.r.t ‘x’ on the number line.
Now, let us calculate LHL and RHL for equation (1);
LHL=$\underset{x\to 0-}{\mathop{\lim }}\,\dfrac{{{[x]}^{2}}}{{{x}^{2}}}$
Hence x is lying between $(-1,0)$ as $x\to {{0}^{-}}$(left to 0) and hence [x] = -1.
Hence, LHL after putting [x] = -1, we get
$\underset{x\to 0-}{\mathop{\lim }}\,\dfrac{{{\left( -1 \right)}^{2}}}{{{x}^{2}}}=\underset{x\to 0-}{\mathop{\lim }}\,\dfrac{1}{{{x}^{2}}}$
As$x\to {{0}^{-}}$, i.e. x maybe (-0.000123), so we will get some definite value of $\dfrac{1}{{{x}^{2}}}$if we apply$x\to {{0}^{-}}$.
Hence, LHL gives a definite value.
Now, RHL can be calculated as
RHL=$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{{{[x]}^{2}}}{{{x}^{2}}}$
As x is lying between (0, 1) for$x\to {{0}^{+}}$,
hence [x] will be 0.
Hence, RHL=$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{0}{{{x}^{2}}}$
Here, we can put $x\to {{0}^{+}}$(example: x=0.000333), but we will get RHL=0 as the numerator is already 0.
Hence, LHL and RHL for $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{[x]}^{2}}}{{{x}^{2}}}=l$will not exist as LHL given a definite value and RHL is giving 0. Both are not equal.
Now, let us calculate limit for $m=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{\left[ x \right]}^{2}}}{{{x}^{2}}}$i.e. equation (2).
LHL can be given as
LHL=$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{{{\left[ x \right]}^{2}}}{{{x}^{2}}}$
As x is lying in (-1, 0), so ${{x}^{2}}$lies in (0, 1) as squaring negative values between -1 and 0, we will get positive values between (0,1). Hence ${{x}^{2}}$will be positive and tending to ${{0}^{+}}$as x is tending to${{0}^{-}}$. Hence [x] =0.
We get, LHL=$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{0}{{{x}^{2}}}=0$
Hence, LHL will be zero for any definite value of x which is tending to zero.
We get RHL=$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{0}{{{x}^{2}}}=0$
RHL can be calculated as
RHL=$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{{{[x]}^{2}}}{{{x}^{2}}}$
As ‘x’ is lying in (0, 1); hence squaring numbers between (0,1) will be more smaller than 1 and will lie in (0,1) as well. Hence $[{{x}^{2}}]$=0.
We get RHL=$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{0}{{{x}^{2}}}=0$
Hence, RHL will be zero for any definite value of x which is tending to ${{0}^{+}}$.
Therefore, $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{\left[ x \right]}^{2}}}{{{x}^{2}}}=m$will exist as LHL and RHL are equal to each other.
Hence, B. is the correct answer i.e. m exists but l does not.

Note: One can get confuse by putting $x\to 0$ to $\dfrac{[{{x}^{2}}]}{{{x}^{2}}}$, as we can get $\dfrac{0}{0}$ if we put limit directly to the function, which is wrong. Hence, we cannot put x=0 to the given function directly in these kinds of problems.
One can apply LHL and RHL by replacing x by (0-h) and (0+h) to the expressions where h will tend to zero. Now, related LHL and RHL for existing limits.
So, we need to use LHL and RHL for determining the limits of these kinds of questions.