Let$f(x)=\dfrac{{{\sin }^{-1}}(1-\left\{ x \right\}).{{\cos }^{-1}}(1-\left\{ x \right\})}{\sqrt{2\left\{ x \right\}}.(1-\left\{ x \right\})},$then find the left hand limit and right hand limits, i.e., $\underset{x\to {{0}_{+}}}{\mathop{\lim }}\,f(x)\text{ and }\underset{x\to {{0}_{-}}}{\mathop{\lim }}\,f(x).\text{ }$
(Where$\left\{ x \right\}$ denotes the fractional part of x)
Answer
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Hint: Find out right hand and left hand limit separately, using the product rule of limit. Convert Fractional part function to Greatest integer function and solve by substituting \[x\] as \[\left( 0+h \right)\] or \[\left( 0-h \right)\].
As per the question we have to find two limits here, i.e., the left hand limit and right hand limits, i.e.,
$\underset{x\to {{0}_{+}}}{\mathop{\lim }}\,f(x)\text{ and }\underset{x\to {{0}_{-}}}{\mathop{\lim }}\,f(x).\text{ }$
Consider the given expression,
$\underset{x\to 0}{\mathop{\lim }}\,f(x)=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{\sin }^{-1}}(1-\left\{ x \right\}).{{\cos }^{-1}}(1-\left\{ x \right\})}{\sqrt{2\left\{ x \right\}}.(1-\left\{ x \right\})}$
Here $\left\{ x \right\}$ denotes the fractional part of x.
We know fractional parts will always be non-negative and fractional parts are greater than or equal to $'0'$ and less than $'1'$ .
As we all know that,
\[x=[x]+\{x\}\]
\[\therefore \{x\}=x-[x]\]
Therefore the given expression can be written as,
\[\underset{x\to 0}{\mathop{\lim }}\,f(x)=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{\sin }^{-1}}(1-(x-[x])).{{\cos }^{-1}}(1-(x-[x]))}{\sqrt{2(x-[x])}.(1-(x-[x]))}...........(i)\]
Now we will find the right hand limit of the given function.
As we have to find the limits of positive side we should substitute $'x'$ as (\[0+h\])
And as \[x\to {{0}^{+}},h\to 0\], so the above equation (i) can be written as,
\[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{\sin }^{-1}}(1-(0+h-[0+h])).{{\cos }^{-1}}(1-(0+h-[0+h]))}{\sqrt{2(0+h-[0+h])}.(1-(0+h-[0+h]))}\]
As \[h\] is approaching to zero, therefore it can be considered as fraction, \[\left[ 0+h \right]=0\] as the value of greatest integer function is an integer with neglecting the fraction, considering these the above equation can be written as,
\[\begin{align}
& \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{\sin }^{-1}}(1-(h-0)).{{\cos }^{-1}}(1-(h-0))}{\sqrt{2(h-0)}.(1-(h-0))} \\
& \Rightarrow \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{\sin }^{-1}}(1-h).{{\cos }^{-1}}(1-h)}{\sqrt{2h}.(1-h)} \\
\end{align}\]
Now we know the limit of a product is the product of the limits. So, the limit of product of two functions is equal to the product of individual limits of the functions, that is, the above function can be written as,
$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{\sin }^{-1}}(1-h)}{(1-h)}.\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{\cos }^{-1}}(1-h)}{\sqrt{2h}}$
Now applying the limits to sine function, we get
$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\dfrac{{{\sin }^{-1}}(1)}{(1)}.\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{\cos }^{-1}}(1-h)}{\sqrt{2h}}$
Now we know ${{\sin }^{-1}}(1)=\dfrac{\pi }{2}$, so the above equation becomes,
$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\dfrac{\pi }{2}.\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{\cos }^{-1}}(1-h)}{\sqrt{2h}}........(i)$
Now if we apply the limits, we see that
$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\dfrac{\pi }{2}.\dfrac{{{\cos }^{-1}}(1-0)}{\sqrt{2(0)}}=\dfrac{0}{0}$
This is indeterminate form, so we will apply L’ HOSPITAL RULE i.e., differentiating numerator and denominator. So, differentiating numerator and denominator of equation (i) with respect to $h$ , we get
$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\dfrac{\pi }{2}.\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{d}{dh}\left( {{\cos }^{-1}}(1-h) \right)}{\dfrac{d}{dh}\left( \sqrt{2h} \right)}$
The differentiation of ${{\cos }^{-1}}x$ is $-\dfrac{1}{\sqrt{1-{{x}^{2}}}}$ , applying this formula, the above equation becomes,
$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\dfrac{\pi }{2}.\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{-1}{\sqrt{1-{{(1-h)}^{2}}}}\dfrac{d}{dh}(1-h)}{\sqrt{2}\times \dfrac{1}{2\sqrt{h}}}$
Solving this, we get
\[\begin{align}
& \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\dfrac{\pi }{2}.\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{-1}{\sqrt{1-{{(1-h)}^{2}}}}(-1)}{\dfrac{1}{\sqrt{2h}}} \\
& \Rightarrow \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\dfrac{\pi }{2}.\underset{h\to 0}{\mathop{\lim }}\,\left[ \dfrac{1}{\sqrt{1-{{(1-h)}^{2}}}}\times \sqrt{2h} \right] \\
& \Rightarrow \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\dfrac{\pi }{2}.\underset{h\to 0}{\mathop{\lim }}\,\left[ \dfrac{1}{\sqrt{1-(1+{{h}^{2}}-2h)}}\times \sqrt{2h} \right] \\
& \Rightarrow \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\dfrac{\pi }{2}.\underset{h\to 0}{\mathop{\lim }}\,\left[ \dfrac{1}{\sqrt{1-1-{{h}^{2}}+2h}}\times \sqrt{2h} \right] \\
& \Rightarrow \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\dfrac{\pi }{2}.\underset{h\to 0}{\mathop{\lim }}\,\left[ \sqrt{\dfrac{2h}{2h-{{h}^{2}}}} \right] \\
\end{align}\]
Taking out the common term, we get
\[\Rightarrow \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\dfrac{\pi }{2}.\underset{h\to 0}{\mathop{\lim }}\,\left[ \sqrt{\dfrac{2h}{2h(1-h)}} \right]\]
Cancelling the like term, we get
\[\Rightarrow \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\dfrac{\pi }{2}.\underset{h\to 0}{\mathop{\lim }}\,\left[ \sqrt{\dfrac{1}{1(1-h)}} \right]\]
Applying the limit, we get
\[\Rightarrow \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\dfrac{\pi }{2}.\left[ \sqrt{\dfrac{1}{1(1-0)}} \right]=\dfrac{\pi }{2}\]
So the right hand limit is $\dfrac{\pi }{2}$ .
Now we will find the left hand limit of the given function.
$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{{{\sin }^{-1}}(1-\left\{ x \right\}).{{\cos }^{-1}}(1-\left\{ x \right\})}{\sqrt{2\left\{ x \right\}}.(1-\left\{ x \right\})}$
Now we know the limit of a product is the product of the limits. So, the limit of product of two functions is equal to the product of individual limits of the functions, that is, the above function can be written as,
$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{{{\sin }^{-1}}(1-\left\{ x \right\})}{(1-\left\{ x \right\})}.\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{{{\cos }^{-1}}(1-\left\{ x \right\})}{\sqrt{2\left\{ x \right\}}}$
Now as $x$ is tending to zero, i.e., $x$ is less than zero, so we will substitute$\left\{ x \right\}=-h$, so as $x\to 0$, therefore $h\to 0$, so the above equation can be written as,
$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{\sin }^{-1}}(1+h)}{(1+h)}.\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{\cos }^{-1}}(1+h)}{\sqrt{2(-h)}}$
Now we know ${{\sin }^{-1}}(1)=\dfrac{\pi }{2}$, so the above equation becomes,
\[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\dfrac{\pi }{2}.\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{\cos }^{-1}}(1+h)}{\sqrt{2(-h)}}........(ii)\]
Now if we apply the limits, we see that
$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\dfrac{{{\cos }^{-1}}(1+0)}{\sqrt{2(-0)}}=\dfrac{0}{0}$
This is indeterminate form, so we will apply L’ HOSPITAL RULE i.e., differentiating numerator and denominator. So, differentiating numerator and denominator of equation (ii) with respect to $h$ , we get
\[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\dfrac{\pi }{2}.\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{d}{dh}\left( {{\cos }^{-1}}(1+h) \right)}{\dfrac{d}{dh}\left( \sqrt{2(-h)} \right)}\]
The differentiation of ${{\cos }^{-1}}x$ is $-\dfrac{1}{\sqrt{1-{{x}^{2}}}}$ , applying this formula, the above equation becomes,
$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\dfrac{\pi }{2}.\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{-1}{\sqrt{1-{{(1+h)}^{2}}}}\dfrac{d}{dh}(1+h)}{\sqrt{-2}\times \dfrac{(1)}{2\sqrt{h}}}$
Solving this, we get
\[\begin{align}
& \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\dfrac{\pi }{2}.\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{-1}{\sqrt{1-{{(1+h)}^{2}}}}(1)}{\dfrac{(1)}{\sqrt{-2h}}} \\
& \Rightarrow \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\dfrac{\pi }{2}.\underset{h\to 0}{\mathop{\lim }}\,\left[ -\dfrac{1}{\sqrt{1-{{(1+h)}^{2}}}}\times \sqrt{-2h} \right] \\
& \Rightarrow \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\dfrac{\pi }{2}.\underset{h\to 0}{\mathop{\lim }}\,\left[ -\dfrac{1}{\sqrt{1-(1+{{h}^{2}}+2h)}}\times \sqrt{-2h} \right] \\
& \Rightarrow \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\dfrac{\pi }{2}.\underset{h\to 0}{\mathop{\lim }}\,\left[ -\dfrac{1}{\sqrt{1-1-{{h}^{2}}-2h}}\times \sqrt{-2h} \right] \\
& \Rightarrow \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\dfrac{\pi }{2}.\underset{h\to 0}{\mathop{\lim }}\,\left[ \sqrt{\dfrac{-2h}{-2h-{{h}^{2}}}} \right] \\
\end{align}\]
Taking out the common term, we get
\[\Rightarrow \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\dfrac{\pi }{2}.\underset{h\to 0}{\mathop{\lim }}\,\left[ -\sqrt{\dfrac{-2h}{-2h(1+h)}} \right]\]
Cancelling the like term, we get
\[\Rightarrow \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\dfrac{\pi }{2}.\underset{h\to 0}{\mathop{\lim }}\,\left[ -\sqrt{\dfrac{1}{1(1+h)}} \right]\]
Applying the limit, we get
\[\Rightarrow \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\dfrac{\pi }{2}.\left[ -\sqrt{\dfrac{1}{1(1-0)}} \right]=-\dfrac{\pi }{2}\]
So the left hand limit is $-\dfrac{\pi }{2}$ .
$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\dfrac{\pi }{2}\text{ and }\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=-\dfrac{\pi }{2}$
Note: The common mistake is substituting $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{\sin }^{-1}}x}{x}=1$, so we get
$\begin{align}
& \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{{{\sin }^{-1}}(1-h)}{(1-h)}.\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{{{\cos }^{-1}}(1-h)}{\sqrt{2h}} \\
& \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=(1).\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{{{\cos }^{-1}}(1-h)}{\sqrt{2h}} \\
\end{align}$
In this case we will get a completely different answer.
As per the question we have to find two limits here, i.e., the left hand limit and right hand limits, i.e.,
$\underset{x\to {{0}_{+}}}{\mathop{\lim }}\,f(x)\text{ and }\underset{x\to {{0}_{-}}}{\mathop{\lim }}\,f(x).\text{ }$
Consider the given expression,
$\underset{x\to 0}{\mathop{\lim }}\,f(x)=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{\sin }^{-1}}(1-\left\{ x \right\}).{{\cos }^{-1}}(1-\left\{ x \right\})}{\sqrt{2\left\{ x \right\}}.(1-\left\{ x \right\})}$
Here $\left\{ x \right\}$ denotes the fractional part of x.
We know fractional parts will always be non-negative and fractional parts are greater than or equal to $'0'$ and less than $'1'$ .
As we all know that,
\[x=[x]+\{x\}\]
\[\therefore \{x\}=x-[x]\]
Therefore the given expression can be written as,
\[\underset{x\to 0}{\mathop{\lim }}\,f(x)=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{\sin }^{-1}}(1-(x-[x])).{{\cos }^{-1}}(1-(x-[x]))}{\sqrt{2(x-[x])}.(1-(x-[x]))}...........(i)\]
Now we will find the right hand limit of the given function.
As we have to find the limits of positive side we should substitute $'x'$ as (\[0+h\])
And as \[x\to {{0}^{+}},h\to 0\], so the above equation (i) can be written as,
\[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{\sin }^{-1}}(1-(0+h-[0+h])).{{\cos }^{-1}}(1-(0+h-[0+h]))}{\sqrt{2(0+h-[0+h])}.(1-(0+h-[0+h]))}\]
As \[h\] is approaching to zero, therefore it can be considered as fraction, \[\left[ 0+h \right]=0\] as the value of greatest integer function is an integer with neglecting the fraction, considering these the above equation can be written as,
\[\begin{align}
& \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{\sin }^{-1}}(1-(h-0)).{{\cos }^{-1}}(1-(h-0))}{\sqrt{2(h-0)}.(1-(h-0))} \\
& \Rightarrow \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{\sin }^{-1}}(1-h).{{\cos }^{-1}}(1-h)}{\sqrt{2h}.(1-h)} \\
\end{align}\]
Now we know the limit of a product is the product of the limits. So, the limit of product of two functions is equal to the product of individual limits of the functions, that is, the above function can be written as,
$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{\sin }^{-1}}(1-h)}{(1-h)}.\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{\cos }^{-1}}(1-h)}{\sqrt{2h}}$
Now applying the limits to sine function, we get
$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\dfrac{{{\sin }^{-1}}(1)}{(1)}.\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{\cos }^{-1}}(1-h)}{\sqrt{2h}}$
Now we know ${{\sin }^{-1}}(1)=\dfrac{\pi }{2}$, so the above equation becomes,
$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\dfrac{\pi }{2}.\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{\cos }^{-1}}(1-h)}{\sqrt{2h}}........(i)$
Now if we apply the limits, we see that
$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\dfrac{\pi }{2}.\dfrac{{{\cos }^{-1}}(1-0)}{\sqrt{2(0)}}=\dfrac{0}{0}$
This is indeterminate form, so we will apply L’ HOSPITAL RULE i.e., differentiating numerator and denominator. So, differentiating numerator and denominator of equation (i) with respect to $h$ , we get
$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\dfrac{\pi }{2}.\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{d}{dh}\left( {{\cos }^{-1}}(1-h) \right)}{\dfrac{d}{dh}\left( \sqrt{2h} \right)}$
The differentiation of ${{\cos }^{-1}}x$ is $-\dfrac{1}{\sqrt{1-{{x}^{2}}}}$ , applying this formula, the above equation becomes,
$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\dfrac{\pi }{2}.\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{-1}{\sqrt{1-{{(1-h)}^{2}}}}\dfrac{d}{dh}(1-h)}{\sqrt{2}\times \dfrac{1}{2\sqrt{h}}}$
Solving this, we get
\[\begin{align}
& \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\dfrac{\pi }{2}.\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{-1}{\sqrt{1-{{(1-h)}^{2}}}}(-1)}{\dfrac{1}{\sqrt{2h}}} \\
& \Rightarrow \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\dfrac{\pi }{2}.\underset{h\to 0}{\mathop{\lim }}\,\left[ \dfrac{1}{\sqrt{1-{{(1-h)}^{2}}}}\times \sqrt{2h} \right] \\
& \Rightarrow \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\dfrac{\pi }{2}.\underset{h\to 0}{\mathop{\lim }}\,\left[ \dfrac{1}{\sqrt{1-(1+{{h}^{2}}-2h)}}\times \sqrt{2h} \right] \\
& \Rightarrow \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\dfrac{\pi }{2}.\underset{h\to 0}{\mathop{\lim }}\,\left[ \dfrac{1}{\sqrt{1-1-{{h}^{2}}+2h}}\times \sqrt{2h} \right] \\
& \Rightarrow \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\dfrac{\pi }{2}.\underset{h\to 0}{\mathop{\lim }}\,\left[ \sqrt{\dfrac{2h}{2h-{{h}^{2}}}} \right] \\
\end{align}\]
Taking out the common term, we get
\[\Rightarrow \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\dfrac{\pi }{2}.\underset{h\to 0}{\mathop{\lim }}\,\left[ \sqrt{\dfrac{2h}{2h(1-h)}} \right]\]
Cancelling the like term, we get
\[\Rightarrow \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\dfrac{\pi }{2}.\underset{h\to 0}{\mathop{\lim }}\,\left[ \sqrt{\dfrac{1}{1(1-h)}} \right]\]
Applying the limit, we get
\[\Rightarrow \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\dfrac{\pi }{2}.\left[ \sqrt{\dfrac{1}{1(1-0)}} \right]=\dfrac{\pi }{2}\]
So the right hand limit is $\dfrac{\pi }{2}$ .
Now we will find the left hand limit of the given function.
$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{{{\sin }^{-1}}(1-\left\{ x \right\}).{{\cos }^{-1}}(1-\left\{ x \right\})}{\sqrt{2\left\{ x \right\}}.(1-\left\{ x \right\})}$
Now we know the limit of a product is the product of the limits. So, the limit of product of two functions is equal to the product of individual limits of the functions, that is, the above function can be written as,
$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{{{\sin }^{-1}}(1-\left\{ x \right\})}{(1-\left\{ x \right\})}.\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{{{\cos }^{-1}}(1-\left\{ x \right\})}{\sqrt{2\left\{ x \right\}}}$
Now as $x$ is tending to zero, i.e., $x$ is less than zero, so we will substitute$\left\{ x \right\}=-h$, so as $x\to 0$, therefore $h\to 0$, so the above equation can be written as,
$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{\sin }^{-1}}(1+h)}{(1+h)}.\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{\cos }^{-1}}(1+h)}{\sqrt{2(-h)}}$
Now we know ${{\sin }^{-1}}(1)=\dfrac{\pi }{2}$, so the above equation becomes,
\[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\dfrac{\pi }{2}.\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{\cos }^{-1}}(1+h)}{\sqrt{2(-h)}}........(ii)\]
Now if we apply the limits, we see that
$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\dfrac{{{\cos }^{-1}}(1+0)}{\sqrt{2(-0)}}=\dfrac{0}{0}$
This is indeterminate form, so we will apply L’ HOSPITAL RULE i.e., differentiating numerator and denominator. So, differentiating numerator and denominator of equation (ii) with respect to $h$ , we get
\[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\dfrac{\pi }{2}.\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{d}{dh}\left( {{\cos }^{-1}}(1+h) \right)}{\dfrac{d}{dh}\left( \sqrt{2(-h)} \right)}\]
The differentiation of ${{\cos }^{-1}}x$ is $-\dfrac{1}{\sqrt{1-{{x}^{2}}}}$ , applying this formula, the above equation becomes,
$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\dfrac{\pi }{2}.\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{-1}{\sqrt{1-{{(1+h)}^{2}}}}\dfrac{d}{dh}(1+h)}{\sqrt{-2}\times \dfrac{(1)}{2\sqrt{h}}}$
Solving this, we get
\[\begin{align}
& \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\dfrac{\pi }{2}.\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{-1}{\sqrt{1-{{(1+h)}^{2}}}}(1)}{\dfrac{(1)}{\sqrt{-2h}}} \\
& \Rightarrow \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\dfrac{\pi }{2}.\underset{h\to 0}{\mathop{\lim }}\,\left[ -\dfrac{1}{\sqrt{1-{{(1+h)}^{2}}}}\times \sqrt{-2h} \right] \\
& \Rightarrow \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\dfrac{\pi }{2}.\underset{h\to 0}{\mathop{\lim }}\,\left[ -\dfrac{1}{\sqrt{1-(1+{{h}^{2}}+2h)}}\times \sqrt{-2h} \right] \\
& \Rightarrow \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\dfrac{\pi }{2}.\underset{h\to 0}{\mathop{\lim }}\,\left[ -\dfrac{1}{\sqrt{1-1-{{h}^{2}}-2h}}\times \sqrt{-2h} \right] \\
& \Rightarrow \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\dfrac{\pi }{2}.\underset{h\to 0}{\mathop{\lim }}\,\left[ \sqrt{\dfrac{-2h}{-2h-{{h}^{2}}}} \right] \\
\end{align}\]
Taking out the common term, we get
\[\Rightarrow \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\dfrac{\pi }{2}.\underset{h\to 0}{\mathop{\lim }}\,\left[ -\sqrt{\dfrac{-2h}{-2h(1+h)}} \right]\]
Cancelling the like term, we get
\[\Rightarrow \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\dfrac{\pi }{2}.\underset{h\to 0}{\mathop{\lim }}\,\left[ -\sqrt{\dfrac{1}{1(1+h)}} \right]\]
Applying the limit, we get
\[\Rightarrow \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\dfrac{\pi }{2}.\left[ -\sqrt{\dfrac{1}{1(1-0)}} \right]=-\dfrac{\pi }{2}\]
So the left hand limit is $-\dfrac{\pi }{2}$ .
$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\dfrac{\pi }{2}\text{ and }\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=-\dfrac{\pi }{2}$
Note: The common mistake is substituting $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{\sin }^{-1}}x}{x}=1$, so we get
$\begin{align}
& \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{{{\sin }^{-1}}(1-h)}{(1-h)}.\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{{{\cos }^{-1}}(1-h)}{\sqrt{2h}} \\
& \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=(1).\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{{{\cos }^{-1}}(1-h)}{\sqrt{2h}} \\
\end{align}$
In this case we will get a completely different answer.
Last updated date: 27th Sep 2023
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