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Question: How do you find the limit of \(\dfrac{{\sin x}}{{2{x^2} - x}}\) as \(x\) approaches \(0\) ?...

How do you find the limit of sin⁡x2x2−x\dfrac{{\sin x}}{{2{x^2} - x}} as xx approaches 00 ?

Explanation

Solution

To find the limits of any function, we have to put values in the function, after applying limit if the function is in the determinate form or we will get finite value then we will proceed further and get answer otherwise if the function is in the indeterminate form as 00\dfrac{0}{0} or ∞∞\dfrac{\infty }{\infty } then we will use L‘Hospital’s Rule and then get value of function in determinate or finite form and then proceed further to get answer.
Formula/concept used:
L'hospital's rule –
When any function after applying limits gives you a value of function as indeterminate form or infinite then we have to differentiate numerator and denominator separately to obtain a finite or determinable value.

Complete step by step solution: We have given function lim⁡x→0sin⁡(x)2x2−x\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin \left( x \right)}}{{2{x^2} - x}}
Now, we put limit value in function which is x→0x \to 0 , so we get:
⇒sin⁡(0)2(0)2−0=00\Rightarrow \dfrac{{\sin \left( 0 \right)}}{{2{{\left( 0 \right)}^2} - 0}} = \dfrac{0}{0}
We obtain resultant after putting limit it is in indeterminate form i.e. 00\dfrac{0}{0}
We will apply l ‘Hospital’s Rule we will differentiate parts of fraction numerator and denominator separately, we obtain:
=lim⁡x→0ddx(sin⁡(x))(2x2−x)= \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{d}{{dx}}\left( {\sin \left( x \right)} \right)}}{{\left( {2{x^2} - x} \right)}}
=lim⁡x→0cos⁡(x)4x2−1= \mathop {\lim }\limits_{x \to 0} \dfrac{{\cos \left( x \right)}}{{4{x^2} - 1}}
We know that
\left\\{ {\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x{\text{ & }}\dfrac{d}{{dx}}\left( {{x^2}} \right) = 2x{\text{ & }}\dfrac{d}{{dx}}\left( x \right) = 1} \right\\}
Now,
Substituting variable of the function with the value of limit given, we get:
⇒lim⁡x→0cos⁡(x)4x2−1\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\cos \left( x \right)}}{{4{x^2} - 1}}
On simplifying we get =cos⁡(0)4(0)−1=10−1 = \dfrac{{\cos \left( 0 \right)}}{{4\left( 0 \right) - 1}} = \dfrac{1}{{0 - 1}}
We get =1−1 = \dfrac{1}{{ - 1}}
=−1= - 1
Therefore, we get limit of given function sin⁡x2x2−x\dfrac{{\sin x}}{{2{x^2} - x}} as xx approaches 00is
lim⁡x→0sin⁡(x)2x2−x=−1\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin \left( x \right)}}{{2{x^2} - x}} = - 1

Note:
After applying L'hospital's Rule to the function if we again get 00\dfrac{0}{0} or ∞∞\dfrac{\infty }{\infty } form, then again apply the L hospital’s Rule. So, stop applying the rule when you have a determinable or finite form of the given function.
If we do not get the determinable form then we have to apply Cauchy mean value theorem for Taylor series.