Try this beautiful problem based on Limit, useful for ISI B.Stat Entrance

## Problem on Limit | ISI B.Stat TOMATO 728

The limit lim \(\int\frac {h}{(h^2 + x^2)}\)dx (integration running from \(x =-1\)to \(x = 1\)) as\( h \to 0\)

- equals 0
- equals \(\pi\)
- equals \(-\pi\)
- deoes not exist

**Key Concepts**

Limit

Calculas

trigonometry

## Check the Answer

But try the problem first…

Answer: does not exist

TOMATO, Problem 728

Challenges and Thrills in Pre College Mathematics

## Try with Hints

First hint

Now, \(\int{h}{(h^2 + x^2)}\)dx (integration running from \(x = -1\) to \(x = 1\))

Let, \(x\) = h tany

\(\Rightarrow dx = h sec^2y dy\)

\(\Rightarrow \) \(x = -1\), \(y = -tan^{-1}(1/h)\) and \(x = 1\), \(y = tan^{-1}(1/h)\)

\(\Rightarrow \int \frac{h}{(h^2 + x^2)}\)dx =\(\int \frac{h(hsec^2ydy)}{h^2sec^2y}\) (integration running from\( y = -tan^{-1}(1/h) \) to \(y = tan^-1(1/h))\)

= y (upper limit =\( tan-1(1/h)\)) and lower limit = \(-tan^-1(1/h)\)

= \(2tan^-1(1/h)\)

Can you now finish the problem ……….

Second Hint

Now, lim \(2tan^-(1/h)\) as\( h \to 0\) doesn‟t exist

## Other useful links

- https://www.cheenta.com/sum-of-co-ordinates-amc-10a-2014-problem-21/
- https://www.youtube.com/watch?v=fRj9NuPGrLU&t=276s

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