Question

Let $\underset{x\to 0}{lim}{\sec }^{-1}\left({\displaystyle \frac{x}{\sin x}}\right)=l$ and $$\underset{x\to 0}{lim}{\sec }^{-1}\left({\displaystyle \frac{x}{\tan x}}\right)=m$$, 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 1

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
$\underset{x\to 0}{lim}{\sec }^{-1}\left({\displaystyle \frac{x}{\sin x}}\right)=l....\left(i\right)$
And, $$\underset{x\to 0}{lim}{\sec }^{-1}\left({\displaystyle \frac{x}{\tan x}}\right)=m.....\left(ii\right)$$
First, we need to know about the domain of $${\sec }^{-1}x$$ i.e. $$(-\mathrm{\infty },-1]\cup [1,\mathrm{\infty })$$.
Now, try to relate values of $${\displaystyle \frac{x}{\sin x}}$$ and $${\displaystyle \frac{x}{\tan x}}$$ for limit $$x\to 0$$, if value inside of $${\sec }^{-1}\left(\right)$$ will lie in $$(-1,1)$$ then limit will not exist and if value inside the bracket lies in $$(-\mathrm{\infty },-1]\cup [1,\mathrm{\infty })$$. Hence limit will exist.
Let us first relate $${\displaystyle \frac{x}{\sin x}}$$.
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 $$(x_1,y_1)$$ and curve is y = f (x) then tangent at $$(x_1,y_1)$$ can be given by $$y-y_1={{\displaystyle \frac{dy}{dx}}|}_{(x_1,y_1)}(x-x_1)$$
Tangent equation for sin x at (0, 0) is
$$y-0={{\displaystyle \frac{d}{dx}}(\sin x)|}_{(0,0)}(x-0)$$
$$y={\cos x|}_{(0,0)}\left(x\right)[\because {\displaystyle \frac{d}{dx}}\sin x=\cos x]$$
$$y=x$$
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. $${\displaystyle \frac{x}{\tan x}}$$, we get the tangent equation of tan x at (0, 0) is
$$y-0={{\displaystyle \frac{dy}{dx}}|}_{(0,0)}(x-0)$$
$$y={{\sec }^{2}x|}_{(0,0)}\left(x\right)[{\displaystyle \frac{d}{dx}}\tan x={\sec }^{2}x]$$
$$y=x$$
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 $${\displaystyle \frac{x}{\sin x}}$$ that is:
Case 1: $$x\to 0^{+}$$
We observe x > sin x
Hence, $${\displaystyle \frac{x}{\sin x}}>1$$
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
$$x>\sin x$$
$${\displaystyle \frac{x}{\sin x}}>1$$
Hence, from case 1 and case 2, we get
If $$limx\to 0,$$ then $${\displaystyle \frac{x}{\sin x}}>1....\left(iii\right)$$
Similarly, let us relate x and tan x for $$x\to 0$$
Case 1: $$x\to 0^{+}$$
x < tan x
$${\displaystyle \frac{x}{\tan x}}<1$$
Case 2: $$x\to 0^{-}$$
x < tan x
$${\displaystyle \frac{x}{\tan x}}<1$$
Hence, for $$x\to 0$$, we have $${\displaystyle \frac{x}{\tan x}}<1....\left(iv\right)$$
Now, for limit ‘l’ from equation (i), we get
$$l=\underset{x\to 0}{lim}{\sec }^{-1}\left({\displaystyle \frac{x}{\sin x}}\right)$$
As we have $${\displaystyle \frac{x}{\sin x}}>1$$ from equation (iii) and domain of $${\sec }^{-1}x$$ is $$(-\mathrm{\infty },-1]\cup [1,\mathrm{\infty })$$ as explained in the starting. Hence, we can put $$limx\to 0$$ to the given relation.
So, $$l=\underset{x\to 0}{lim}{\sec }^{-1}\left({\displaystyle \frac{x}{\sin x}}\right)$$ will exist.
For limit ‘m’ from equation (ii), we get
$$m=\underset{x\to 0}{lim}{\sec }^{-1}\left({\displaystyle \frac{x}{\tan x}}\right)$$
We have already calculated that $${\displaystyle \frac{x}{\tan x}}<1$$ from equation (iv) and domain of $${\sec }^{-1}x$$ is $$(-\mathrm{\infty },-1]\cup [1,\mathrm{\infty })$$. Hence the given limit will not exist.
Hence, option (a) is the correct answer to the given problem.

Note: One can directly put $$\underset{x\to 0}{lim}{\displaystyle \frac{x}{\sin x}}=1$$ and $$\underset{x\to 0}{lim}{\displaystyle \frac{x}{\tan x}}=1$$ as we generally use but that will be wrong for the given expression. As the exact value of $$\underset{x\to 0}{lim}{\displaystyle \frac{x}{\sin x}}$$ and $$\underset{x\to 0}{lim}{\displaystyle \frac{x}{\tan x}}$$ 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.