# Evaluate the limit \[\underset{n\to \infty }{\mathop{\lim }}\,\left[ \dfrac{\sqrt{n}}{{{(3+4\sqrt{n})}^{2}}}+\dfrac{\sqrt{n}}{\sqrt{2}{{(3\sqrt{2}+4\sqrt{n})}^{2}}}+\dfrac{\sqrt{n}}{\sqrt{3}{{(3\sqrt{3}+4\sqrt{n})}^{2}}}....+\dfrac{1}{49n} \right]\]

Answer

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Hint: Try to convert the sum into integral using the idea that integral is a limit of a sum.

We can see that the \[{{k}^{th}}\] term of the limit can be written as\[\dfrac{\sqrt{n}}{\sqrt{k}{{(3\sqrt{k}+4\sqrt{n})}^{2}}}\]

Therefore,

\[\underset{n\to \infty }{\mathop{\lim }}\,\left[ \dfrac{\sqrt{n}}{{{(3+4\sqrt{n})}^{2}}}+\dfrac{\sqrt{n}}{\sqrt{2}{{(3\sqrt{2}+4\sqrt{n})}^{2}}}+\dfrac{\sqrt{n}}{\sqrt{3}{{(3\sqrt{3}+4\sqrt{n})}^{2}}}....+\dfrac{1}{49n} \right]\]

=\[\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{k=1}^{n}{\dfrac{\sqrt{n}}{\sqrt{k}{{(3\sqrt{k}+4\sqrt{n})}^{2}}}}\]

The limits of the summation can be found by equating the \[{{k}^{th}}\] term to the first term (for the lower limit) and to the last term (for the upper limit), which in our case are \[k=1\] and \[k=n\].

Dividing by \[\dfrac{1}{\sqrt{n}}\] from both numerator and denominator we rewrite our equation as,

= \[\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{k=1}^{n}{\dfrac{1}{n}\dfrac{1}{\sqrt{\dfrac{k}{n}}{{\left( 3\sqrt{\dfrac{k}{n}}+4 \right)}^{2}}}}\]

This limit can now be replaced by an integral as \[\dfrac{1}{n}=dx\] and \[\dfrac{k}{n}=x\].

The upper and lower limits of the integral can be found by evaluating the limit of \[\dfrac{k}{n}\]as \[k\] tends to upper and lower bounds of our summation, which in our case are \[0\] (as lower bound) and \[1\] as upper bound. Every limit of the form

\[\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{k=an}^{bn}{\dfrac{1}{n}f\left( \dfrac{k}{n} \right)}\] can be converted to the integral \[\int\limits_{a}^{b}{f(x)}dx\].

So,

= \[\int\limits_{0}^{1}{\dfrac{1}{\sqrt{x}{{\left( 3\sqrt{x}+4 \right)}^{2}}}}dx\]

For evaluating this integral substitute\[t=\sqrt{x}\]. So \[dt=\dfrac{1}{2\sqrt{x}}dx\].Upper bound becomes \[1\] and lower bound becomes \[0\] (these remain the same in our case as the substitution was \[t=\sqrt{x}\], but may change if it would have been something else)

= \[\int\limits_{0}^{1}{\dfrac{2}{{{\left( 3t+4 \right)}^{2}}}}dt\]

= \[\left[ \dfrac{-2}{3\left( 3t+4 \right)} \right]_{0}^{1}\]

= \[\left( \dfrac{-2}{21} \right)-\left( \dfrac{-2}{12} \right)\]

=\[\dfrac{1}{14}\]

Note: Students have to be careful when writing the \[{{k}^{th}}\] term of the summation and its limits. Also, when converting the limit to the integral one should be careful. Students can use their own methods i.e. any other substitution when solving the integral. They may use either the lower Reimann sum or the upper Reimann sum, in both cases the answer will be the same.

We can see that the \[{{k}^{th}}\] term of the limit can be written as\[\dfrac{\sqrt{n}}{\sqrt{k}{{(3\sqrt{k}+4\sqrt{n})}^{2}}}\]

Therefore,

\[\underset{n\to \infty }{\mathop{\lim }}\,\left[ \dfrac{\sqrt{n}}{{{(3+4\sqrt{n})}^{2}}}+\dfrac{\sqrt{n}}{\sqrt{2}{{(3\sqrt{2}+4\sqrt{n})}^{2}}}+\dfrac{\sqrt{n}}{\sqrt{3}{{(3\sqrt{3}+4\sqrt{n})}^{2}}}....+\dfrac{1}{49n} \right]\]

=\[\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{k=1}^{n}{\dfrac{\sqrt{n}}{\sqrt{k}{{(3\sqrt{k}+4\sqrt{n})}^{2}}}}\]

The limits of the summation can be found by equating the \[{{k}^{th}}\] term to the first term (for the lower limit) and to the last term (for the upper limit), which in our case are \[k=1\] and \[k=n\].

Dividing by \[\dfrac{1}{\sqrt{n}}\] from both numerator and denominator we rewrite our equation as,

= \[\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{k=1}^{n}{\dfrac{1}{n}\dfrac{1}{\sqrt{\dfrac{k}{n}}{{\left( 3\sqrt{\dfrac{k}{n}}+4 \right)}^{2}}}}\]

This limit can now be replaced by an integral as \[\dfrac{1}{n}=dx\] and \[\dfrac{k}{n}=x\].

The upper and lower limits of the integral can be found by evaluating the limit of \[\dfrac{k}{n}\]as \[k\] tends to upper and lower bounds of our summation, which in our case are \[0\] (as lower bound) and \[1\] as upper bound. Every limit of the form

\[\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{k=an}^{bn}{\dfrac{1}{n}f\left( \dfrac{k}{n} \right)}\] can be converted to the integral \[\int\limits_{a}^{b}{f(x)}dx\].

So,

= \[\int\limits_{0}^{1}{\dfrac{1}{\sqrt{x}{{\left( 3\sqrt{x}+4 \right)}^{2}}}}dx\]

For evaluating this integral substitute\[t=\sqrt{x}\]. So \[dt=\dfrac{1}{2\sqrt{x}}dx\].Upper bound becomes \[1\] and lower bound becomes \[0\] (these remain the same in our case as the substitution was \[t=\sqrt{x}\], but may change if it would have been something else)

= \[\int\limits_{0}^{1}{\dfrac{2}{{{\left( 3t+4 \right)}^{2}}}}dt\]

= \[\left[ \dfrac{-2}{3\left( 3t+4 \right)} \right]_{0}^{1}\]

= \[\left( \dfrac{-2}{21} \right)-\left( \dfrac{-2}{12} \right)\]

=\[\dfrac{1}{14}\]

Note: Students have to be careful when writing the \[{{k}^{th}}\] term of the summation and its limits. Also, when converting the limit to the integral one should be careful. Students can use their own methods i.e. any other substitution when solving the integral. They may use either the lower Reimann sum or the upper Reimann sum, in both cases the answer will be the same.

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