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Evaluate the limit $\underset{x\to \dfrac{\pi }{4}}{\mathop{\lim }}\,\dfrac{\mathop{\int }_{2}^{{{\sec }^{2}}x}f\left( t \right)dt}{{{x}^{2}}-\dfrac{{{\pi }^{2}}}{16}}$
A) $\dfrac{8}{\pi }f(2)$
B) $\dfrac{2}{\pi }f(2)$
C) $\dfrac{2}{\pi }f\left( \dfrac{1}{2} \right)$
D) $2f(2)$
Answer
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Hint: See that the limit is of the form $\dfrac{0}{0}$ and there is an integral in limit. Use Leibniz Rule and L’Hopital Rule to evaluate the limit.
Let the given limit be equal to \[L\],
$L=\underset{x\to \dfrac{\pi }{4}}{\mathop{\lim }}\,\dfrac{\mathop{\int }_{2}^{{{\sec }^{2}}x}f\left( t \right)dt}{{{x}^{2}}-\dfrac{{{\pi }^{2}}}{16}}$
If we substitute \[x\] as $\dfrac{\pi }{4}$ we get,
$L=\dfrac{\mathop{\int }_{2}^{{{\sec }^{2}}\dfrac{\pi }{4}}f\left( t \right)dt}{{{\left( \dfrac{\pi }{4} \right)}^{2}}-\dfrac{{{\pi }^{2}}}{16}}$
\[L=\dfrac{\mathop{\int }_{2}^{{{\sqrt{2}}^{2}}}f\left( t \right)dt}{\dfrac{{{\pi }^{2}}}{16}-\dfrac{{{\pi }^{2}}}{16}}\]
\[L=\dfrac{\mathop{\int }_{2}^{2}f\left( t \right)dt}{\dfrac{{{\pi }^{2}}}{16}-\dfrac{{{\pi }^{2}}}{16}}\]
We know that \[\int\limits_{a}^{a}{f(x)dx=0}\]. Therefore, the numerator tends to \[0\].
\[L=\dfrac{0}{0}\]
The numerator and denominator tends to zero as \[x\] tends to $\dfrac{\pi }{4}$, so the limit is of the form $\dfrac{0}{0}$. So we can use L’Hopital’s Rule i.e. differentiating the numerator and denominator separately to evaluate the limit. L’Hopital Rule can only be used when the limit is of the form $\dfrac{0}{0}$ or$\dfrac{\infty }{\infty }$.
$L=\underset{x\to \dfrac{\pi }{4}}{\mathop{\lim }}\,\dfrac{\dfrac{d}{dx}\left( \mathop{\int }_{2}^{{{\sec }^{2}}x}f\left( t \right)dt \right)}{\dfrac{d}{dx}\left( {{x}^{2}}-\dfrac{{{\pi }^{2}}}{16} \right)}$
For evaluating the numerator we use Leibniz’s Rule i.e.
$\dfrac{d}{dx}\left( \mathop{\int }_{b\left( x \right)}^{a\left( x \right)}f\left( x \right)dx \right)=\left\{ f\left( a\left( x \right) \right)~a'\left( x \right) \right\}-\left\{ f\left( b\left( x \right) \right)b'\left( x \right) \right\}$ where\[a'\left( x \right)\]and \[b'\left( x \right)\]are derivatives of functions \[a\left( x \right)\]and \[b\left( x \right)\] with respect to\[x\].
$\dfrac{d}{dx}\left( \mathop{\int }_{2}^{{{\sec }^{2}}x}f\left( t \right)dt \right)=f\left( se{{c}^{2}}\left( x \right) \right)~\left( 2se{{c}^{2}}\left( x \right)\tan \left( x \right) \right)-\left( f\left( 2 \right)\left( 0 \right) \right)$
$\dfrac{d}{dx}\left( \mathop{\int }_{2}^{{{\sec }^{2}}x}f\left( t \right)dt \right)=2f\left( se{{c}^{2}}\left( x \right) \right)se{{c}^{2}}\left( x \right))\text{tan}\left( x \right)$
So,
$L=\underset{x\to \dfrac{\pi }{4}}{\mathop{\lim }}\,\dfrac{2f\left( se{{c}^{2}}\left( x \right) \right)se{{c}^{2}}\left( x \right))\text{tan}\left( x \right)}{2x}$
Now we can simply evaluate the limit by substituting \[x\] as$\dfrac{\pi }{4}$.
$\text{L= }\!\!~\!\!\text{ }\dfrac{2f\left( se{{c}^{2}}\left( \text{ }\!\!~\!\!\text{ }\dfrac{\pi }{4} \right) \right)se{{c}^{2}}\left( \text{ }\!\!~\!\!\text{ }\dfrac{\pi }{4} \right))\text{tan}\left( \text{ }\!\!~\!\!\text{ }\dfrac{\pi }{4} \right)}{2~\times ~\text{ }\!\!~\!\!\text{ }\dfrac{\pi }{4}}$
$\text{L= }\!\!~\!\!\text{ }\dfrac{2f\left( 2 \right)\times 2\times 1}{\dfrac{\pi }{2}}$
$\text{L= }\!\!~\!\!\text{ }\dfrac{8}{\pi }f\left( 2 \right)$
So, the answer is Option A) $\dfrac{8}{\pi }f(2)$
Note: Students must be careful while using Leibniz Rule and L'Hopital Rule. They might make mistakes by only differentiating the numerator only the denominator only, or not using Leibniz Rule correctly i.e. they might not differentiate the limits, not put the limits correctly, etc. Do not use L'Hopital's Rule multiple times, it may lead to incorrect answers.
Let the given limit be equal to \[L\],
$L=\underset{x\to \dfrac{\pi }{4}}{\mathop{\lim }}\,\dfrac{\mathop{\int }_{2}^{{{\sec }^{2}}x}f\left( t \right)dt}{{{x}^{2}}-\dfrac{{{\pi }^{2}}}{16}}$
If we substitute \[x\] as $\dfrac{\pi }{4}$ we get,
$L=\dfrac{\mathop{\int }_{2}^{{{\sec }^{2}}\dfrac{\pi }{4}}f\left( t \right)dt}{{{\left( \dfrac{\pi }{4} \right)}^{2}}-\dfrac{{{\pi }^{2}}}{16}}$
\[L=\dfrac{\mathop{\int }_{2}^{{{\sqrt{2}}^{2}}}f\left( t \right)dt}{\dfrac{{{\pi }^{2}}}{16}-\dfrac{{{\pi }^{2}}}{16}}\]
\[L=\dfrac{\mathop{\int }_{2}^{2}f\left( t \right)dt}{\dfrac{{{\pi }^{2}}}{16}-\dfrac{{{\pi }^{2}}}{16}}\]
We know that \[\int\limits_{a}^{a}{f(x)dx=0}\]. Therefore, the numerator tends to \[0\].
\[L=\dfrac{0}{0}\]
The numerator and denominator tends to zero as \[x\] tends to $\dfrac{\pi }{4}$, so the limit is of the form $\dfrac{0}{0}$. So we can use L’Hopital’s Rule i.e. differentiating the numerator and denominator separately to evaluate the limit. L’Hopital Rule can only be used when the limit is of the form $\dfrac{0}{0}$ or$\dfrac{\infty }{\infty }$.
$L=\underset{x\to \dfrac{\pi }{4}}{\mathop{\lim }}\,\dfrac{\dfrac{d}{dx}\left( \mathop{\int }_{2}^{{{\sec }^{2}}x}f\left( t \right)dt \right)}{\dfrac{d}{dx}\left( {{x}^{2}}-\dfrac{{{\pi }^{2}}}{16} \right)}$
For evaluating the numerator we use Leibniz’s Rule i.e.
$\dfrac{d}{dx}\left( \mathop{\int }_{b\left( x \right)}^{a\left( x \right)}f\left( x \right)dx \right)=\left\{ f\left( a\left( x \right) \right)~a'\left( x \right) \right\}-\left\{ f\left( b\left( x \right) \right)b'\left( x \right) \right\}$ where\[a'\left( x \right)\]and \[b'\left( x \right)\]are derivatives of functions \[a\left( x \right)\]and \[b\left( x \right)\] with respect to\[x\].
$\dfrac{d}{dx}\left( \mathop{\int }_{2}^{{{\sec }^{2}}x}f\left( t \right)dt \right)=f\left( se{{c}^{2}}\left( x \right) \right)~\left( 2se{{c}^{2}}\left( x \right)\tan \left( x \right) \right)-\left( f\left( 2 \right)\left( 0 \right) \right)$
$\dfrac{d}{dx}\left( \mathop{\int }_{2}^{{{\sec }^{2}}x}f\left( t \right)dt \right)=2f\left( se{{c}^{2}}\left( x \right) \right)se{{c}^{2}}\left( x \right))\text{tan}\left( x \right)$
So,
$L=\underset{x\to \dfrac{\pi }{4}}{\mathop{\lim }}\,\dfrac{2f\left( se{{c}^{2}}\left( x \right) \right)se{{c}^{2}}\left( x \right))\text{tan}\left( x \right)}{2x}$
Now we can simply evaluate the limit by substituting \[x\] as$\dfrac{\pi }{4}$.
$\text{L= }\!\!~\!\!\text{ }\dfrac{2f\left( se{{c}^{2}}\left( \text{ }\!\!~\!\!\text{ }\dfrac{\pi }{4} \right) \right)se{{c}^{2}}\left( \text{ }\!\!~\!\!\text{ }\dfrac{\pi }{4} \right))\text{tan}\left( \text{ }\!\!~\!\!\text{ }\dfrac{\pi }{4} \right)}{2~\times ~\text{ }\!\!~\!\!\text{ }\dfrac{\pi }{4}}$
$\text{L= }\!\!~\!\!\text{ }\dfrac{2f\left( 2 \right)\times 2\times 1}{\dfrac{\pi }{2}}$
$\text{L= }\!\!~\!\!\text{ }\dfrac{8}{\pi }f\left( 2 \right)$
So, the answer is Option A) $\dfrac{8}{\pi }f(2)$
Note: Students must be careful while using Leibniz Rule and L'Hopital Rule. They might make mistakes by only differentiating the numerator only the denominator only, or not using Leibniz Rule correctly i.e. they might not differentiate the limits, not put the limits correctly, etc. Do not use L'Hopital's Rule multiple times, it may lead to incorrect answers.
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