Question

Find the value of $\mathop {\lim }\limits_{\theta \to \dfrac{\pi }{2}} \dfrac{{1 - \sin \theta }}{{\left( {\dfrac{\pi }{2} - \theta } \right)\cos \theta }}$.
A. $1$
B. $ - 1$
C. $\dfrac{1}{2}$
D. $ - \dfrac{1}{2}$

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 $\mathop {\lim }\limits_{\theta \to \dfrac{\pi }{2}} \dfrac{{1 - \sin \theta }}{{\left( {\dfrac{\pi }{2} - \theta } \right)\cos \theta }}$ using L’Hospital’s rule. So, if we put the limit x tending to $\left( {\dfrac{\pi }{2}} \right)$ into the expression $\dfrac{{1 - \sin \theta }}{{\left( {\dfrac{\pi }{2} - \theta } \right)\cos \theta }}$, 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 $\theta $ separately and then apply the limit.
Hence, $\mathop {\lim }\limits_{\theta \to \dfrac{\pi }{2}} \dfrac{{1 - \sin \theta }}{{\left( {\dfrac{\pi }{2} - \theta } \right)\cos \theta }}$
Now, the derivative of $\sin \theta $ is $\cos \theta $ and the derivative of $\cos \theta $ is $\sin \theta $ . Also, we must know the product rule of differentiation $\dfrac{{d\left[ {f\left( x \right) \times g\left( x \right)} \right]}}{{dx}} = f\left( x \right)g'\left( x \right) + g\left( x \right)f'\left( x \right)$ to find derivative of the denominator. So, we get,
$\mathop {\lim }\limits_{\theta \to \dfrac{\pi }{2}} \dfrac{{ - \cos \theta }}{{\left( {\dfrac{\pi }{2} - \theta } \right)\left( { - \sin \theta } \right) + \cos \theta \left( { - 1} \right)}}$

Now, simplifying the expression, we get,
$\mathop {\lim }\limits_{\theta \to \dfrac{\pi }{2}} \dfrac{{ - \cos \theta }}{{\left( {\theta - \dfrac{\pi }{2}} \right)\sin \theta - \cos \theta }}$
But still if we substitute the limit x tending to $\left( {\dfrac{\pi }{2}} \right)$ into the expression $\dfrac{{ - \cos \theta }}{{\left( {\theta - \dfrac{\pi }{2}} \right)\sin \theta - \cos \theta }}$, we get an indeterminate form limit. So, we apply the L’Hospital’s rule one more time. Differentiating the numerator and denominator with respect to $\theta $, we get,
$\mathop {\lim }\limits_{\theta \to \dfrac{\pi }{2}} \dfrac{{ - \left( { - \sin \theta } \right)}}{{\left( {\theta - \dfrac{\pi }{2}} \right)\left( {\cos \theta } \right) + \sin \theta \left( 1 \right) - \left( { - \sin \theta } \right)}}$

Simplifying the expression, we get,
$\mathop {\lim }\limits_{\theta \to \dfrac{\pi }{2}} \dfrac{{\sin \theta }}{{\left( {\theta - \dfrac{\pi }{2}} \right)\left( {\cos \theta } \right) + 2\sin \theta }}$
Now substituting the value of limit x tending to $\left( {\dfrac{\pi }{2}} \right)$ into the expression, we get,
$\dfrac{{\sin \left( {\dfrac{\pi }{2}} \right)}}{{\left( {\dfrac{\pi }{2} - \dfrac{\pi }{2}} \right)\left( {\cos \dfrac{\pi }{2}} \right) + 2\sin \left( {\dfrac{\pi }{2}} \right)}}$
Now, we know that the value of $\sin \dfrac{\pi }{2}$ is $1$ and $\cos \dfrac{\pi }{2}$ is $0$. So, we get,
$\dfrac{1}{{0 \times 0 + 2\left( 1 \right)}}$
$\therefore \dfrac{1}{2}$
So, the value of the limit $\mathop {\lim }\limits_{\theta \to \dfrac{\pi }{2}} \dfrac{{1 - \sin \theta }}{{\left( {\dfrac{\pi }{2} - \theta } \right)\cos \theta }}$ is $\left( {\dfrac{1}{2}} \right)$.

Hence, option C is the correct answer.

Note: Always check before evaluating the problem whether it is of indeterminate or determinate form. Remember the L’Hospital’s rule 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.