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# What is the limit of $\dfrac{{\sin \left( {2x} \right)}}{{{x^2}}}$ as $x$ approaches 0? Verified
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Hint: Here the question in the form of word, so first we write in the actual form as $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin \left( {2x} \right)}}{{{x^2}}}$ and then we will apply the limit to the function. When we apply the limit to the function, after applying the limit if we get the real number then we write as it is, otherwise if we get an undetermined form then we apply the L’ Hospital’s rule.

Historically, Sir Issac Newton was the "inventor" of derivatives and Leibnitz introduced the concept of Limits. Limits and derivatives are extremely crucial concepts in Maths whose application is not only limited to Maths but are also present in other subjects like physics.
A limit is defined as a value that a function approaches as the input, and it produces some value. Limits are important in calculus and mathematical analysis and used to define integrals, derivatives, and continuity.
To express the limit of a function, we represent it as: $\mathop {\lim }\limits_{n \to c} f\left( n \right) = L$
Now consider the given question and it is written as
$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin \left( {2x} \right)}}{{{x^2}}}$

On applying the limit to the function, we have
$\Rightarrow \dfrac{{\sin \left( {2\left( 0 \right)} \right)}}{{{{\left( 0 \right)}^2}}}$
On simplifying this we get
$\Rightarrow \dfrac{0}{0}$
This is an undetermined form so we solve this by applying the L’ Hospital’s rule.
In mathematics, more specifically calculus, L’ Hospital’s rule provides a technique to evaluate limits of indeterminate forms. It is denoted as.
$\mathop {\lim }\limits_{x \to c} \dfrac{{f'(x)}}{{g'(x)}}$
Here, $f\left( x \right) = \sin 2x$ and $g\left( x \right) = {x^2}$
On applying the L’ Hospital’s rule for the given limit function we have
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{d}{{dx}}\left( {\sin \left( {2x} \right)} \right)}}{{\dfrac{d}{{dx}}\left( {{x^2}} \right)}}$
On differentiating, we have
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{2\cos \left( {2x} \right)}}{{2x}}$
On simplifying the above inequality, we have
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\cos \left( {2x} \right)}}{x}$
Now applying the limit that is as $x \to 0$ , we have
$\Rightarrow \dfrac{{\cos \left( {2\left( 0 \right)} \right)}}{0}$
On simplifying this we get
$\Rightarrow \dfrac{0}{0}$
This is also undetermined form so again by applying the L’ Hospital’s rule, then
$\mathop {\lim }\limits_{x \to c} \dfrac{{f''(x)}}{{g''(x)}}$
Here, $f''(x) = \cos \left( {2x} \right)$ and $g''(x) = x$ , then we have
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{d}{{dx}}\left( {\cos \left( {2x} \right)} \right)}}{{\dfrac{d}{{dx}}\left( x \right)}}$
On differentiating, we have
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{ - 2\sin \left( {2x} \right)}}{1}$
On simplifying the above inequality, we have
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \left( { - 2\sin \left( {2x} \right)} \right)$
Now applying the limit that is as $x \to 0$ , we have
$\Rightarrow - 2\sin \left( {2\left( 0 \right)} \right)$
$\Rightarrow - 2\sin \left( 0 \right)$
$\Rightarrow 0$
Because the value of $\sin \left( 0 \right) = 0$ , then
$\therefore \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin \left( {2x} \right)}}{{{x^2}}} = 0$
Hence, it’s a required solution.
So, the correct answer is “0”.

Note: Whenever we get the limit value of the function in the indeterminate form it is necessary to apply the L’ Hospital’s rule for the given function. While applying the L’ Hospital’s rule we have to differentiate each function with respect to x and again we apply the limit for the function which is differentiated.