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For the following limit check whether LHL and RHL exist or not:$\underset{x\to 8}{\mathop{\lim }}\,\dfrac{\sin \left\{ x-10 \right\}}{\left\{ 10-x \right\}}$ (where { } denotes fractional part function)(a) LHL exist but RHL does not exist (b) RHL exist but LHL does not exist(c) Neither LHL nor RHL does not exist(d) Both RHL and LHL exist and equals to 1

Last updated date: 13th Jul 2024
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Hint: Use the basic definition of fractional part function with respect to greatest integer i.e. {x} = x – [x].

Here, we have the limit given as $\underset{x\to 8}{\mathop{\lim }}\,\dfrac{\sin \left\{ x-10 \right\}}{\left\{ 10-x \right\}}$ where { } is denoting fractional function.
Let us suppose the given limit is ‘L’.
Hence, we have
$L=\underset{x\to 8}{\mathop{\lim }}\,\dfrac{\sin \left\{ x-10 \right\}}{\left\{ 10-x \right\}}....\left( i \right)$
Now, we have to find the limit of the given expression. Hence we need to find the LHL (Left-hand limit) and RHL (Right-hand limit) of the given expression.
Therefore, for LHL, we have to put $x\to {{8}^{-}}$. Hence we get
$L=\underset{x\to {{8}^{-}}}{\mathop{\lim }}\,\dfrac{\sin \left\{ x-10 \right\}}{\left\{ 10-x \right\}}$
Now, replacing ‘x’ by ( 8 – h) where $h\to 0$, we get
$L=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin \left\{ 8-h-10 \right\}}{\left\{ 10-\left( 8-h \right) \right\}}$
$L=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin \left\{ -2-h \right\}}{\left\{ 2+h \right\}}....\left( ii \right)$
As we know that {x} will lie on [0, 1) and is defined as {x} = x – [x] where [ ] denotes the greatest integer.
Hence, we can put x = 2 + h, we get,
{2 + h} = 2 + h – [2 + h]
As $h\to 0$, hence [2 + h] will open as ‘2’. As (2 + h) will lie between [2, 3) and the greatest integer of that will be 2. Hence we get
$\left\{ 2+h \right\}=2+h-2=h....\left( iii \right)$
Now putting x = – 2 – h in {x} = x – [x], we get
$\left\{ -2-h \right\}=-2-h-\left[ -2-h \right]$
As, (-2, -h) will lie on [-3, -2) and hence greatest integer of – 2 – h will be – 3. Hence, we get
$\left\{ -2-h \right\}=-2-h+3=1-h....\left( iv \right)$
Hence, equation (ii) will be written as $L=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( 1-h \right)}{h}=\dfrac{\sin 1}{0}$ which is not defined and tending to $\infty$. Now, coming to the RHL part, i.e. we need to put $x\to {{8}^{+}}$ or replacing x by (h + 8), where $h\to 0$. Hence, we get
$L=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin \left\{ 8+h-10 \right\}}{\left\{ 10-8-h \right\}}$
$L=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin \left\{ -2+h \right\}}{\left\{ 2-h \right\}}....\left( v \right)$
Now, by relation {x} = x – [x], we get {-2 + h} = – 2 + h – [–2 + h]
Since (–2 + h) will lie in (–2, –1) and hence the greatest integer of (–2 + h) is –2. Therefore,
$\left\{ -2+h \right\}=-2+h+2=h....\left( vi \right)$
Now, for {2 – h}, we get
$\left\{ 2-h \right\}=2-h-\left[ 2-h \right]$
As (2 – h) will lie in (1, 2) and hence the greatest integer will be 1.
Hence, we get
$\left\{ 2-h \right\}=2-h-1=1-h....\left( vii \right)$
Now, putting values of {–2 + h} and {2 – h} from equations (vi) and (vii) in equation (v), we get
$L=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin \text{ }h}{1-h}=0$
Hence, the value of RHL is definite and equal to 0.
Therefore, RHL exists and LHL does not exist for the given limit.
Hence, option (b) is the correct answer.

Note: One can directly put x = 8 to the given relation and get $\dfrac{\sin \left\{ -2 \right\}}{\left\{ 2 \right\}}$ where {- 2} and {2} will be zero. And hence we get an indeterminate form of limit i.e. $\dfrac{0}{0}$. And one can compare this relation by $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x}=1$, and give an answer that limit will exist and equal to ‘1’ which is the wrong approach for these kinds of questions.
Hence, we cannot put direct limits to functions like {x}, [x], |x| or log |x| etc. Therefore, always try to calculate LHL and RHL, both for these kinds of questions.