# Let $\underset{x\to 0}{\mathop{\lim }}\,{{\sec }^{-1}}\left( \dfrac{x}{\sin x} \right)=l$ and \[\underset{x\to 0}{\mathop{\lim }}\,{{\sec }^{-1}}\left( \dfrac{x}{\tan x} \right)=m\], then

(a) l exists but m does not

(b) m exists but l does not

(c) l and m both exist

(d) neither l nor m exists

Answer

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Hint: Relate the relation between x, sin x and tan x when x is limiting to zero. Relate it with the domain of \[{{\sec }^{-1}}x\] for existing limits.

Here, we have given the limits as

$\underset{x\to 0}{\mathop{\lim }}\,{{\sec }^{-1}}\left( \dfrac{x}{\sin x} \right)=l....\left( i \right)$

And, \[\underset{x\to 0}{\mathop{\lim }}\,{{\sec }^{-1}}\left( \dfrac{x}{\tan x} \right)=m.....\left( ii \right)\]

First, we need to know about the domain of \[{{\sec }^{-1}}x\] i.e. \[\left( -\infty ,-1 \right]\cup \left[ 1,\infty \right)\].

Now, try to relate values of \[\dfrac{x}{\sin x}\] and \[\dfrac{x}{\tan x}\] for limit \[x\to 0\], if value inside of \[{{\sec }^{-1}}\left( {} \right)\] will lie in \[\left( -1,1 \right)\] then limit will not exist and if value inside the bracket lies in \[\left( -\infty ,-1 \right]\cup \left[ 1,\infty \right)\]. Hence limit will exist.

Let us first relate \[\dfrac{x}{\sin x}\].

One can relate x with sin x and tan x by calculating tangent equations of tan x and sin x at (0, 0) and relate it with y = x.

We know that one can find tangent at any point lying on the curve by calculating slope at that point. Let the point be \[\left( {{x}_{1}},{{y}_{1}} \right)\] and curve is y = f (x) then tangent at \[\left( {{x}_{1}},{{y}_{1}} \right)\] can be given by \[y-{{y}_{1}}={{\left. \dfrac{dy}{dx} \right|}_{\left( {{x}_{1}},{{y}_{1}} \right)}}\left( x-{{x}_{1}} \right)\]

Tangent equation for sin x at (0, 0) is

\[y-0={{\left. \dfrac{d}{dx}\left( \sin x \right) \right|}_{\left( 0,0 \right)}}\left( x-0 \right)\]

\[y={{\left. \cos x \right|}_{\left( 0,0 \right)}}\left( x \right)\text{ }\left[ \because \dfrac{d}{dx}\sin x=\cos x \right]\]

\[y=x\]

Hence, \[y=x\] is tangent for \[y=\sin x\].

Draw graph of x and sin x in one coordinate plane as follows:

Now for the second case i.e. \[\dfrac{x}{\tan x}\], we get the tangent equation of tan x at (0, 0) is

\[y-0={{\left. \dfrac{dy}{dx} \right|}_{\left( 0,0 \right)}}\left( x-0 \right)\]

\[y={{\left. {{\sec }^{2}}x \right|}_{\left( 0,0 \right)}}\left( x \right)\text{ }\left[ \dfrac{d}{dx}\tan x={{\sec }^{2}}x \right]\]

\[y=x\]

Hence, y = x is tangent for \[y=\tan x\] as well.

Let us draw the graph of x and tan x as follows:

Now from the graphs, we can relate for \[\dfrac{x}{\sin x}\] that is:

Case 1: \[x\to {{0}^{+}}\]

We observe x > sin x

Hence, \[\dfrac{x}{\sin x}>1\]

Case 2: \[x\to {{0}^{-}}\]

Here, sin x has a higher positive magnitude than x. Hence, if we put a negative sign to both x and sin x, then

\[x>\sin x\]

\[\dfrac{x}{\sin x}>1\]

Hence, from case 1 and case 2, we get

If \[\lim x\to 0,\] then \[\dfrac{x}{\sin x}>1....\left( iii \right)\]

Similarly, let us relate x and tan x for \[x\to 0\]

Case 1: \[x\to {{0}^{+}}\]

x < tan x

\[\dfrac{x}{\tan x}<1\]

Case 2: \[x\to {{0}^{-}}\]

x < tan x

\[\dfrac{x}{\tan x}<1\]

Hence, for \[x\to 0\], we have \[\dfrac{x}{\tan x}<1....\left( iv \right)\]

Now, for limit ‘l’ from equation (i), we get

\[l=\underset{x\to 0}{\mathop{\lim }}\,{{\sec }^{-1}}\left( \dfrac{x}{\sin x} \right)\]

As we have \[\dfrac{x}{\sin x}>1\] from equation (iii) and domain of \[{{\sec }^{-1}}x\] is \[\left( -\infty ,-1 \right]\cup \left[ 1,\infty \right)\] as explained in the starting. Hence, we can put \[\lim x\to 0\] to the given relation.

So, \[l=\underset{x\to 0}{\mathop{\lim }}\,{{\sec }^{-1}}\left( \dfrac{x}{\sin x} \right)\] will exist.

For limit ‘m’ from equation (ii), we get

\[m=\underset{x\to 0}{\mathop{\lim }}\,{{\sec }^{-1}}\left( \dfrac{x}{\tan x} \right)\]

We have already calculated that \[\dfrac{x}{\tan x}<1\] from equation (iv) and domain of \[{{\sec }^{-1}}x\] is \[\left( -\infty ,-1 \right]\cup \left[ 1,\infty \right)\]. Hence the given limit will not exist.

Hence, option (a) is the correct answer to the given problem.

Note: One can directly put \[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{x}{\sin x}=1\] and \[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{x}{\tan x}=1\] as we generally use but that will be wrong for the given expression. As the exact value of \[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{x}{\sin x}\] and \[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{x}{\tan x}\] is not exactly 1, it’s the limiting value of the given expressions. Hence, be careful with these kinds of problems. Relating x with tan x and sin x by calculating tangent at (0, 0) for sin x and tan x is the key point of the question.

Here, we have given the limits as

$\underset{x\to 0}{\mathop{\lim }}\,{{\sec }^{-1}}\left( \dfrac{x}{\sin x} \right)=l....\left( i \right)$

And, \[\underset{x\to 0}{\mathop{\lim }}\,{{\sec }^{-1}}\left( \dfrac{x}{\tan x} \right)=m.....\left( ii \right)\]

First, we need to know about the domain of \[{{\sec }^{-1}}x\] i.e. \[\left( -\infty ,-1 \right]\cup \left[ 1,\infty \right)\].

Now, try to relate values of \[\dfrac{x}{\sin x}\] and \[\dfrac{x}{\tan x}\] for limit \[x\to 0\], if value inside of \[{{\sec }^{-1}}\left( {} \right)\] will lie in \[\left( -1,1 \right)\] then limit will not exist and if value inside the bracket lies in \[\left( -\infty ,-1 \right]\cup \left[ 1,\infty \right)\]. Hence limit will exist.

Let us first relate \[\dfrac{x}{\sin x}\].

One can relate x with sin x and tan x by calculating tangent equations of tan x and sin x at (0, 0) and relate it with y = x.

We know that one can find tangent at any point lying on the curve by calculating slope at that point. Let the point be \[\left( {{x}_{1}},{{y}_{1}} \right)\] and curve is y = f (x) then tangent at \[\left( {{x}_{1}},{{y}_{1}} \right)\] can be given by \[y-{{y}_{1}}={{\left. \dfrac{dy}{dx} \right|}_{\left( {{x}_{1}},{{y}_{1}} \right)}}\left( x-{{x}_{1}} \right)\]

Tangent equation for sin x at (0, 0) is

\[y-0={{\left. \dfrac{d}{dx}\left( \sin x \right) \right|}_{\left( 0,0 \right)}}\left( x-0 \right)\]

\[y={{\left. \cos x \right|}_{\left( 0,0 \right)}}\left( x \right)\text{ }\left[ \because \dfrac{d}{dx}\sin x=\cos x \right]\]

\[y=x\]

Hence, \[y=x\] is tangent for \[y=\sin x\].

Draw graph of x and sin x in one coordinate plane as follows:

Now for the second case i.e. \[\dfrac{x}{\tan x}\], we get the tangent equation of tan x at (0, 0) is

\[y-0={{\left. \dfrac{dy}{dx} \right|}_{\left( 0,0 \right)}}\left( x-0 \right)\]

\[y={{\left. {{\sec }^{2}}x \right|}_{\left( 0,0 \right)}}\left( x \right)\text{ }\left[ \dfrac{d}{dx}\tan x={{\sec }^{2}}x \right]\]

\[y=x\]

Hence, y = x is tangent for \[y=\tan x\] as well.

Let us draw the graph of x and tan x as follows:

Case 1: \[x\to {{0}^{+}}\]

We observe x > sin x

Hence, \[\dfrac{x}{\sin x}>1\]

Case 2: \[x\to {{0}^{-}}\]

Here, sin x has a higher positive magnitude than x. Hence, if we put a negative sign to both x and sin x, then

\[x>\sin x\]

\[\dfrac{x}{\sin x}>1\]

Hence, from case 1 and case 2, we get

If \[\lim x\to 0,\] then \[\dfrac{x}{\sin x}>1....\left( iii \right)\]

Similarly, let us relate x and tan x for \[x\to 0\]

Case 1: \[x\to {{0}^{+}}\]

x < tan x

\[\dfrac{x}{\tan x}<1\]

Case 2: \[x\to {{0}^{-}}\]

x < tan x

\[\dfrac{x}{\tan x}<1\]

Hence, for \[x\to 0\], we have \[\dfrac{x}{\tan x}<1....\left( iv \right)\]

Now, for limit ‘l’ from equation (i), we get

\[l=\underset{x\to 0}{\mathop{\lim }}\,{{\sec }^{-1}}\left( \dfrac{x}{\sin x} \right)\]

As we have \[\dfrac{x}{\sin x}>1\] from equation (iii) and domain of \[{{\sec }^{-1}}x\] is \[\left( -\infty ,-1 \right]\cup \left[ 1,\infty \right)\] as explained in the starting. Hence, we can put \[\lim x\to 0\] to the given relation.

So, \[l=\underset{x\to 0}{\mathop{\lim }}\,{{\sec }^{-1}}\left( \dfrac{x}{\sin x} \right)\] will exist.

For limit ‘m’ from equation (ii), we get

\[m=\underset{x\to 0}{\mathop{\lim }}\,{{\sec }^{-1}}\left( \dfrac{x}{\tan x} \right)\]

We have already calculated that \[\dfrac{x}{\tan x}<1\] from equation (iv) and domain of \[{{\sec }^{-1}}x\] is \[\left( -\infty ,-1 \right]\cup \left[ 1,\infty \right)\]. Hence the given limit will not exist.

Hence, option (a) is the correct answer to the given problem.

Note: One can directly put \[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{x}{\sin x}=1\] and \[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{x}{\tan x}=1\] as we generally use but that will be wrong for the given expression. As the exact value of \[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{x}{\sin x}\] and \[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{x}{\tan x}\] is not exactly 1, it’s the limiting value of the given expressions. Hence, be careful with these kinds of problems. Relating x with tan x and sin x by calculating tangent at (0, 0) for sin x and tan x is the key point of the question.

Last updated date: 21st Sep 2023

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