Question

Let $f\left( \beta \right) = \mathop {\lim }\limits_{\alpha \to \beta } \dfrac{{{{\sin }^2}\alpha - {{\sin }^2}\beta }}{{{\alpha ^2} - {\beta ^2}}}$ , then $f\left( {\dfrac{\pi }{4}} \right)$ is greater than-
A. $\mathop {\lim }\limits_{x \to \infty } \dfrac{{1 - {{\cos }^3}x}}{{x\sin 2x}}$
B. $\mathop {\lim }\limits_{x \to \infty } \dfrac{{\cot x - \cos x}}{{{{\left( {\pi - 2x} \right)}^3}}}$
C. $\mathop {\lim }\limits_{x \to \infty } \left( {\cos \sqrt {x + 1} - \cos \sqrt x } \right)$
D. $\mathop {\lim }\limits_{x \to a} \dfrac{{\sqrt {a + 2x} - \sqrt {3x} }}{{\sqrt {3a + x} - 2\sqrt x }}$ where $a > 0$

Answer 1

To solve the question given above, we will be using the L’hospital rule. This rule is used to perform differentiation in limits. According to L’hospital rule, $\mathop {\lim }\limits_{x \to \alpha } \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to \alpha } \dfrac{{f'\left( x \right)}}{{g'\left( x \right)}}$ . To solve the above question and questions similar to it, this rule is the most important. You also need to remember the differentiation of trigonometric functions.

Formula used:
L’hospital rule, $\mathop {\lim }\limits_{x \to \alpha } \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to \alpha } \dfrac{{f'\left( x \right)}}{{g'\left( x \right)}}$ .
$\left( {\cos a - \cos b = - 2\sin \dfrac{{a + b}}{2}\sin \dfrac{{a - b}}{2}} \right)$

Complete step by step solution:
We are given: $f\left( \beta \right) = \mathop {\lim }\limits_{x \to \infty } \dfrac{{{{\sin }^2}\alpha - {{\sin }^2}\beta }}{{{\alpha ^2} - {\beta ^2}}}$
Now we can see that this limit is of $\dfrac{0}{0}$ form. So, to solve this question, we will use the L’hospital rule.
According to L’hospital rule, $\mathop {\lim }\limits_{x \to \alpha } \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to \alpha } \dfrac{{f'\left( x \right)}}{{g'\left( x \right)}}$ . Using this, we get,
$\mathop {\lim }\limits_{\alpha \to \beta } \dfrac{{2\sin \alpha \cos \alpha - 0}}{{2\alpha - 0}}$
$\Rightarrow \mathop {\lim }\limits_{\alpha \to \beta } \dfrac{{\sin \alpha \cos \alpha }}{\alpha }$ .
On solving the limit, we get: $\dfrac{{\sin \beta \cos \beta }}{\beta }$
Now, we have to find the value of $f\left( {\dfrac{\pi }{4}} \right)$ .
Put $\beta = \dfrac{\pi }{4}$ ,
$\dfrac{{\sin \dfrac{\pi }{4}\cos \dfrac{\pi }{4}}}{{\dfrac{\pi }{4}}}$ ,
In the next step we will equate the values $\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$ and $\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$ .
We get,

$$\dfrac{{\dfrac{1}{{\sqrt 2 }} \times \dfrac{1}{{\sqrt 2 }}}}{{\dfrac{\pi }{4}}} \\\ \Rightarrow \dfrac{1}{2} \times \dfrac{4}{\pi } \\\$$

$\Rightarrow \dfrac{2}{\pi }$ .
Now, we have to find the option from which this value will be greater.
First, we will find the exact value of $\dfrac{2}{\pi }$ . We get, $\dfrac{2}{\pi } = 0.636$ .
Now, we will check for the first option.
$\mathop {\lim }\limits_{x \to \infty } \dfrac{{1 - {{\cos }^3}x}}{{x\sin 2x}}$
This is again of $\dfrac{0}{0}$ form. So, we will apply L’hospital rule.
We get,
$\mathop {\lim }\limits_{x \to 0} \dfrac{{3{{\cos }^2}x\sin x}}{{2x\cos 2x + \sin 2x}}$
This is again of $\dfrac{0}{0}$ form. So, we will again apply the L'hospital rule.
$\mathop {\lim }\limits_{x \to 0} \dfrac{{3{{\cos }^2}x\cos x + 6{{\cos }^2}x\left( { - \sin x} \right)}}{{2\cos 2x + 2\cos 2x - \left( {4\sin 2x} \right)x}}$
On evaluating the limit, we get,

\dfrac{3}{{2 + 2}} \\\ \Rightarrow \dfrac{3}{4} \\\ \Rightarrow 0.75 \\\ $$ . Now this value is greater than $$f\left( {\dfrac{\pi }{4}} \right)$$ . So, this option is incorrect. Now, we will check for option B. $$\mathop {\lim }\limits_{x \to \infty } \dfrac{{\cot x - \cos x}}{{{{\left( {\pi - 2x} \right)}^3}}}$$ which is of the form $$\dfrac{0}{0}$$ . so, we will differentiate using L’hospital rule. $$\mathop {\lim }\limits_{x \to \dfrac{\pi }{2}} \dfrac{{ - \cos e{c^2}x + \sin x}}{{3{{\left( {\pi - 2x} \right)}^2}\left( { - 2} \right)}}$$ It is of again $$\dfrac{0}{0}$$ form. Using L’hospital rule,

\dfrac{1}{6}\mathop {\lim }\limits_{x \to \dfrac{\pi }{2}} \dfrac{{2\cos ecx\left( { - \cos ecx\cot x} \right) + \cos x}}{{2\left( {\pi - 2x} \right)\left( { - 2} \right)}} \\
\Rightarrow \dfrac{1}{{ - 24}}\mathop {\lim }\limits_{x \to \dfrac{\pi }{2}} \dfrac{{ - 2\cos e{c^2}x\cot x + \cos x}}{{\left( {\pi - 2x} \right)}} \\

It is of again $$\dfrac{0}{0}$$ form. Therefore, using L’hospital rule, we get, $$\dfrac{1}{{ - 24}}\mathop {\lim }\limits_{x \to \dfrac{\pi }{2}} \dfrac{{ - 2\left( {2\cos ecx\left( { - \cos ecx\cot x} \right) + \left( { - \cos e{c^2}x} \right)\left( {\cos e{c^2}x} \right) - \sin x} \right)}}{{ - 2}}$$ On evaluating the limit, we get,

\dfrac{{ - 1}}{{24}} \times \dfrac{1}{2} \\
\Rightarrow \dfrac{{ - 1}}{{48}} \\

This value is less than $$f\left( {\dfrac{\pi }{4}} \right)$$ . So, this option is correct. Now we will check for option C. $$\mathop {\lim }\limits_{x \to \infty } \left( {\cos \sqrt {x + 1} - \cos \sqrt x } \right)$$ , $$\mathop {\lim }\limits_{x \to \infty } \left( { - 2\sin \left( {\dfrac{{\sqrt {x + 1} + \sqrt x }}{2}} \right)\sin \left( {\dfrac{{\sqrt {x + 1} - \sqrt x }}{2}} \right)} \right)$$ by using $$\left( {\cos a - \cos b = - 2\sin \dfrac{{a + b}}{2}\sin \dfrac{{a - b}}{2}} \right)$$ . Now we will rationalize the equation, $$\mathop {\lim }\limits_{x \to \infty } \left( { - 2\sin \left( {\dfrac{{\sqrt {x + 1} + \sqrt x }}{2} \times \dfrac{{\sqrt {x + 1} - \sqrt x }}{{\sqrt {x + 1} - \sqrt x }}} \right)\sin \left( {\dfrac{{\sqrt {x + 1} - \sqrt x }}{2} \times \dfrac{{\sqrt {x + 1} + \sqrt x }}{{\sqrt {x + 1} + \sqrt x }}} \right)} \right)$$ After rationalizing we get, $$\mathop {\lim }\limits_{x \to \infty } \left( { - 2\sin \left( {\dfrac{{x + 1 - x}}{{2\left( {\sqrt {x + 1} - \sqrt x } \right)}}} \right)\sin \left( {\dfrac{{x + 1 - x}}{{2\sqrt {x + 1} + \sqrt x }}} \right)} \right)$$ On evaluating the limit, we get the answer $$0$$ . So, $$f\left( {\dfrac{\pi }{4}} \right)$$ is greater than $$0$$ . so, this option is also correct. Now, we will check option D. $$\mathop {\lim }\limits_{x \to a} \dfrac{{\sqrt {a + 2x} - \sqrt {3x} }}{{\sqrt {3a + x} - 2\sqrt x }}$$ . It is of the form $$\dfrac{0}{0}$$ . so, we will differentiate using L’hospital rule. $$\mathop {\lim }\limits_{x \to a} \dfrac{{\sqrt {a + 2x} - \sqrt {3x} }}{{\sqrt {3a + x} - 2\sqrt x }}$$ . It is of again $$\dfrac{0}{0}$$ form. Therefore, using L’hospital rule, we get, $$\mathop {\lim }\limits_{x \to a} \dfrac{{\dfrac{1}{2}\dfrac{2}{{\sqrt {a + 2x} }} - \dfrac{1}{2}\dfrac{3}{{\sqrt {3x} }}}}{{\dfrac{1}{{2\sqrt {3a + x} }} - \dfrac{2}{{2\sqrt x }}}}$$ On evaluating the limit, we get,

\Rightarrow \dfrac{1}{{\sqrt 3 }}\dfrac{{\dfrac{{ - 1}}{2}}}{{\dfrac{{ - 3}}{4}}} \\
\Rightarrow \dfrac{1}{{\sqrt 3 }}\dfrac{2}{3} \\
\Rightarrow \dfrac{2}{{3\sqrt 3 }} \\

Now its value is $$0.384$$ . $$f\left( {\dfrac{\pi }{4}} \right)$$ is greater than $$0.384$$ . So, this option is correct. So, the final answer: $$f\left( {\dfrac{\pi }{4}} \right)$$ is greater than B. $$\mathop {\lim }\limits_{x \to \infty } \dfrac{{\cot x - \cos x}}{{{{\left( {\pi - 2x} \right)}^3}}}$$ , C. $$\mathop {\lim }\limits_{x \to \infty } \left( {\cos \sqrt {x + 1} - \cos \sqrt x } \right)$$ and D. $$\mathop {\lim }\limits_{x \to a} \dfrac{{\sqrt {a + 2x} - \sqrt {3x} }}{{\sqrt {3a + x} - 2\sqrt x }}$$ . **Note:** To solve questions similar to the above one, you need to remember the L’hospital rule, which is, $$\mathop {\lim }\limits_{x \to \alpha } \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to \alpha } \dfrac{{f'\left( x \right)}}{{g'\left( x \right)}}$$ . You also need to remember the trigonometric differentiation. Without them the question cannot be solved. Solve the question step by step, checking the answer for each option one by one.