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

Hint: 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.