Solveeit Logo
Login

Question

Question: Evaluate the limit of \(\csc x\) as the value of \(x\) approaches \({0^ + }\)....

Evaluate the limit of csc⁡x\csc x as the value of xx approaches 0+{0^ + }.

Explanation

Solution

We know that we say the limit of f(x)f\left( x \right) is LL as xx approaches aa, i.e. lim⁡x→af(x)=L\mathop {\lim }\limits_{x \to a} f\left( x \right) = L. Also we have to make f(x)f\left( x \right) as close to LL and xx as close to aa from both the sides and not letting it be aa. Also using one of the basic trigonometric identities we can write: csc⁡x=1sin⁡x\csc x = \dfrac{1}{{\sin x}}. So by using the above information and expression we can solve the given question.

Complete step by step answer:
Given, csc⁡x........................(i)\csc x........................\left( i \right). Now on comparing the question and the statement of limitlim⁡x→af(x)=L\mathop {\lim }\limits_{x \to a} f\left( x \right) = L; we can say that a=0+  and  f(x)=csc⁡xa = {0^ + }\;{\text{and}}\;{\text{f}}\left( x \right) = \csc x. So by basic definition we can write the statement: the limit of csc⁡x\csc x as the value of xx approaches 0+{0^ + } as:
L=lim⁡x→0+csc⁡x........................(ii)L = \mathop {\lim }\limits_{x \to {0^ + }} \csc x........................(ii)
Since here f(x)=csc⁡x{\text{f}}\left( x \right) = \csc x it’s difficult to find its limit directly so it’s better to use some trigonometric identities since then it would be easier to simplify and solve the given question.

So we know one of the basic trigonometric identity:
csc⁡x=1sin⁡x......................(iii)\csc x = \dfrac{1}{{\sin x}}......................\left( {iii} \right)
So let’s substitute equation (iii) in (ii).
Such that we can write:
lim⁡x→0+csc⁡x=lim⁡x→0+(1sin⁡x).......................(iv)\mathop {\lim }\limits_{x \to {0^ + }} \csc x = \mathop {\lim }\limits_{x \to {0^ + }} \left( {\dfrac{1}{{\sin x}}} \right).......................\left( {iv} \right)
Now we know the exponential identity of inverse which is:(1x)=x−1\left( {\dfrac{1}{x}} \right) = {x^{ - 1}}

So applying the above identity in (iv) we can write:
lim⁡x→0+csc⁡x=lim⁡x→0+(sin⁡x)−1.......................(v)\mathop {\lim }\limits_{x \to {0^ + }} \csc x = \mathop {\lim }\limits_{x \to {0^ + }} {\left( {\sin x} \right)^{ - 1}}.......................\left( v \right)
Now to solve (vi), we know that xx approaches 0+{0^ + }, sin⁡x\sin x also approaches 00 positively such that:
x→0    ⇒sin⁡x→0x \to 0\;\; \Rightarrow \sin x \to 0
But here in the RHS we have the inverse of sin⁡x\sin x such that we can also take the inverse of the limit, that is we can take the inverse of the number 00.
⇒lim⁡x→0+csc⁡x=lim⁡x→0+(sin⁡x)−1 ⇒lim⁡x→0+csc⁡x=10 ∴lim⁡x→0+csc⁡x=∞\Rightarrow \mathop {\lim }\limits_{x \to {0^ + }} \csc x = \mathop {\lim }\limits_{x \to {0^ + }} {\left( {\sin x} \right)^{ - 1}} \\\ \Rightarrow \mathop {\lim }\limits_{x \to {0^ + }} \csc x = \dfrac{1}{0} \\\ \therefore\mathop {\lim }\limits_{x \to {0^ + }} \csc x = \infty

Therefore the limit of csc⁡x\csc x as the value of xx approaches 0+{0^ + } is ∞\infty .

Note: The above given approach for finding limits for similar problems is preferred due to its easier steps and processes. Also while solving a problem containing limits one should be always thorough with all the limit identities as well as the basic trigonometric conversions.