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Solve the given limit : $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{(1-\cos 2x)}{x}$

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Hint: First apply the limit, and check if it evaluates out to an indeterminate form. We can see that, on substituting for the limiting value, it comes out to be in the form of$\dfrac{0}{0}$. So you can apply L’Hopital rule followed by applying the limit, and doing so, you will get the answer.

Now we have given in the question that $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{(1-\cos 2x)}{x}$.
So now let us apply the limit $\underset{x\to 0}{\mathop{\lim }}\,$ in $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{(1-\cos 2x)}{x}$
So applying limit we get,
$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{(1-\cos 2x)}{x}=\dfrac{0}{0}$
Thus, we end up with $\dfrac{0}{0}$, which is an indeterminate form.
The L’Hopital Rule is used to circumvent the common indeterminate forms $\dfrac{0}{0}$ and $\dfrac{\infty }{\infty }$ when computing limits.
L'Hopital rule is the definitive way to simplify the evaluation of limits. It does not directly evaluate limits, but only simplifies evaluation if used appropriately.

This introduces what is called an indeterminate form (there is another, infinity over infinity, but we don't have to worry about that one for this problem).

The differentiation of the numerator and denominator often simplifies the quotient or converts it to a limit that can be evaluated directly.

When direct substitution yields an indeterminate form, we can use L'Hopital rule,
So for this, we can use L’Hopital rule, that is,
If \[\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)}{g\left( x \right)}=\dfrac{0}{0}or\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)}{g\left( x \right)}=\dfrac{\pm \infty }{\pm \infty }\] then \[\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)}{g\left( x \right)}=\underset{x\to a}{\mathop{\lim }}\,\dfrac{{f}'\left( x \right)}{{g}'\left( x \right)}\]

Note that this is not the same as the Product Rule; rather, it means that the limit of the derivative of $f(x)$ over the derivative of $g(x)$ will be the same as the limit of $f(x)$ over $g(x)$.

So here, we can apply the L'Hopital rule.

Also, here we have to use the chain rule.

The chain rule tells us how to find the derivative of a composite function.
The chain rule states that the derivative of $f(g(x))$ is ${{f}^{'}}(g(x))\times {{g}^{'}}(x)$. In other words, it helps us differentiate composite functions.
So differentiation of $f(g(x))={{f}^{'}}(g(x))\times {{g}^{'}}(x)$,
So applying L’Hopital Rule once, and applying the chain rule of differentiation for the numerator, we get:
$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{(0-(-2\sin 2x))}{1}$
So simplifying the above in a simple manner we get,
$\underset{x\to 0}{\mathop{\lim }}\,2\sin 2x$
So now applying limit we get,
$2\sin (0)=0$,
So we get the solution as,
$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{(1-\cos 2x)}{x}=0$


Note: Properly use L’Hopital rule. You should know when to use it. In short L’Hopital rule is \[\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)}{g\left( x \right)}=\dfrac{0}{0}or\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)}{g\left( x \right)}=\dfrac{\pm \infty }{\pm \infty }\] then \[\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)}{g\left( x \right)}=\underset{x\to a}{\mathop{\lim }}\,\dfrac{{f}'\left( x \right)}{{g}'\left( x \right)}\]. You should be familiar with the chain rule. The differentiation of the numerator and denominator often simplifies the quotient or converts it to a limit that can be evaluated directly.