 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

But try the problem first…

Source

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