Question

How do you find the limit of $ \dfrac{{\sin x}}{{x + \sin x}} $ as $ x \to 0 $ ?

Answer 1

Hint: First we take the given trigonometry $ \dfrac{{\sin x}}{{x + \sin x}} $ .This we write in limit is $ \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{{x + \sin x}} $ . Then we use L’ Hospital’s rule and evaluate the resulting expression at zero. After that we compute the derivative of the numerator and compute the derivative of the denominator. Now the derivative substitute in the given equation. After that we apply the limit, hence we get the solution.

Complete Step by Step Solution:
The given trigonometry is $ \dfrac{{\sin x}}{{x + \sin x}} $ as $ x \to 0 $
This write-in limit is
$ \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{{x + \sin x}} $
We find, $ \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{{x + \sin x}} $ =?
We use L’ Hospital’s rule and evaluate the resulting expression at zero.
Because the expression evaluated at zero, is indeterminate, $ \dfrac{0}{0} $ , the use of L’ Hospital’s rule is warranted.
Compute the derivative of the numerator:
$ \Rightarrow \dfrac{{d(\sin (x))}}{{dx}} = \cos (x) $
Compute the derivative of the denominator:
$ \Rightarrow \dfrac{{d(x + \sin (x))}}{{dx}} = 1 + \cos (x) $
Now derivative of $ \sin x $ and $ 1 + \sin x $ substitute in the given trigonometry
$ \Rightarrow \dfrac{{\cos (x)}}{{1 + \cos (x)}} $
Now we take the limit as $ x \to 0 $
$ \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\cos (x)}}{{1 + \cos (x)}} $
Now apply the limit as $ x \to 0 $
$ \Rightarrow \dfrac{{\cos (0)}}{{1 + \cos (0)}} $
Now, $ \cos (0) = 1 $
$ \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{{x + \sin x}} = \dfrac{1}{{1 + 1}} $
Add denominator
$ \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{{x + \sin x}} = \dfrac{1}{2} $

Hence the limit of $ \dfrac{{\sin x}}{{x + \sin x}} $ as $ x \to 0 $ is $ \dfrac{1}{2} $

Note: L’Hospital’s rule can us calculate a limit that may otherwise be hard or impossible. It says that the limit when we divide one function by another is the same after we take the derivative of each function (with some special conditions shown later). In symbols we can write: $ \mathop {\lim }\limits_{x \to c} \dfrac{{f(x)}}{{g(x)}} = \mathop {\lim }\limits_{x \to c} \dfrac{{f'(x)}}{{g'(x)}} $
Cases: We have already seen a $ \dfrac{0}{0} $ and $ \dfrac{\infty }{\infty } $ example, here are all the indeterminate form forms that L’hopital’s rule may be able to help with:
 $ \dfrac{0}{0},\dfrac{\infty }{\infty },{1^\infty },{0^0},{\infty ^0} $
For a limit approaching $ c $ , the original functions must be differentiable on either side of $ c $ , but not necessarily at $ c $ . Likewise $ g'(x) $ were not equal to zero on either side $ c $ .