Question

How do you evaluate the limit $ \dfrac{{\sin x\cos x}}{x} $ as $ x $ approaches $ 0 $ ?

Answer 1

Hint: Use the simple limit formula for $ \dfrac{{\sin x}}{x} $ as $ x $ approaches $ 0 $ ,which gives the value $ 1 $ .Then it would be left with $ \cos x $ for $ x $ approaches $ 0 $ .Just put the value of $ x = 0 $ in $ \cos x $ ,which will give the value as 1.Check the approaches value before applying with the formulas.

Complete step-by-step answer:
Write the given question $ \dfrac{{\sin x\cos x}}{x} $ as $ x $ approaches $ 0 $ , in the basic limit form which is $ \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x\cos x}}{x} $ .
Now, simplify and solve the Question accordingly.
 $ \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x\cos x}}{x} $
It can be written as $ \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} \times \cos x $
Since, we know that limit part would also be separated:
So, $ \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} \times \cos x = \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} \times \mathop {\lim }\limits_{x \to 0} \cos x $
From the basic formulas of limit we know that $ \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1 $
Put the value of $ \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1 $ ,further to simplify.
Hence, we get $ \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} \times \mathop {\lim }\limits_{x \to 0} \cos x = 1 \times \mathop {\lim }\limits_{x \to 0} \cos x $
Now we are left with $ \mathop {\lim }\limits_{x \to 0} \cos x $ .
Since, It cannot be further simplified now, just put $ x = 0 $ in $ \mathop {\lim }\limits_{x \to 0} \cos x $ .
=> $ \mathop {\lim }\limits_{x \to 0} \cos x = \cos (0) $
Since, we know that $ \cos (0) $ =1 therefore, the result obtained is equal to 1.
Hence, $ \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x\cos x}}{x} = 1 $
The limit $ \dfrac{{\sin x\cos x}}{x} $ as $ x $ approaches $ 0 $ gives $ 1 $ .

Note: 1.Always remember the basic formulas of limits for algebraic and trigonometric values, without which the limit question would be very difficult to solve.
2.Always check the approaches value before putting or using the formulas of limits ,as we have seen there was $ x $ approaches $ 0 $ ,but if it was $ 1 $ instead of $ 0 $ ,then the answer would have been wrong and cannot be solved further.
3.Before putting the values to $ x $ always check that the question can further be simplified or not. If it cannot be further simplified then put the value otherwise if it can be further simplified, then simplify and put value.