Answer

Verified

472.5k+ views

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.

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.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Mark and label the given geoinformation on the outline class 11 social science CBSE

When people say No pun intended what does that mea class 8 english CBSE

Name the states which share their boundary with Indias class 9 social science CBSE

Give an account of the Northern Plains of India class 9 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Trending doubts

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

At which age domestication of animals started A Neolithic class 11 social science CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Summary of the poem Where the Mind is Without Fear class 8 english CBSE

One cusec is equal to how many liters class 8 maths CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Change the following sentences into negative and interrogative class 10 english CBSE