Question

How do you find $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 2x}}{{\log \left( {x + 1} \right)}}$ using L’Hospital’s rule?

Answer 1

Hint: The given question requires us to evaluate a limit. A limit is the value that a function (or sequence) approaches as the input (or index) approaches some value. There are various methods and steps to evaluate a limit. Some of the common steps while solving limits involve rationalization and applying some basic results on frequently used limits. L’Hospital’s rule involves solving limits of indeterminate form by differentiating both numerator and denominator with respect to the variable separately and then applying the required limit.

Complete step by step answer:
We have to evaluate limit $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 2x}}{{\log \left( {x + 1} \right)}}$ using L’Hospital’s rule. So, if we put the limit x tending to $0$ into the expression $\dfrac{{\sin 2x}}{{\log \left( {x + 1} \right)}}$, we get an indeterminate form limit. Hence, L’Hospital’s rule can be applied here to find the value of the concerned limit.So, Applying L’Hospital’s rule, we have to differentiate both numerator and denominator with respect to the variable x separately and then apply the limit.Hence,
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 2x}}{{\log \left( {x + 1} \right)}}$

Now, the derivative of \[\sin x\] is $\cos x$ and the derivative of $\log x$ with respect to x is $\dfrac{1}{x}$ . Also, we must know the chain rule of differentiation \[\dfrac{{d\left[ {f\left( {g\left( x \right)} \right)} \right]}}{{dx}} = f'\left( {g\left( x \right)} \right) \times g'\left( x \right)\] to find derivative of the denominator. So, we get,
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {\cos 2x} \right) \times \left( 2 \right)}}{{\dfrac{1}{{\left( {x + 1} \right)}} \times 1}}$
Now, simplifying the expression, we get,
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{2\cos 2x}}{{\dfrac{1}{{\left( {x + 1} \right)}}}}$
$\Rightarrow \mathop {\lim }\limits_{x \to 0} 2\left( {x + 1} \right)\cos 2x$
Taking the constants out of the limit, we get,
$\Rightarrow 2\mathop {\lim }\limits_{x \to 0} \left( {x + 1} \right)\cos 2x$

Now, we substitute the value of the variable as the limit is no longer an indeterminate limit.Hence, we get,
$\Rightarrow 2\left( {0 + 1} \right)\cos \left( {2 \times 0} \right)$
Simplifying the expression, we get,
$\Rightarrow 2\left( 1 \right)\cos \left( 0 \right)$
Now, we know that the cosine of angle zero degrees is one. So, we get,
$\Rightarrow 2\left( 1 \right)\left( 1 \right) = 2$

Therefore, the value of the limit $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 2x}}{{\log \left( {x + 1} \right)}}$ is $2$.

Note: Always check before evaluating the problem whether it is of indeterminate or determinate form. Remember the L’Hospital’s rule is only for evaluating the indeterminate forms. If the limit is of determinate form, then we can substitute the value of limit directly into the expression and get to the final answer. If the limit is of indeterminate form, we use the L’Hospital rule to convert it into determinate form and then substitute the limit.