Try this beautiful problem based on Limit, useful for ISI B.Stat Entrance
The limit lim \(\int\frac {h}{(h^2 + x^2)}\)dx (integration running from \(x =-1\)to \(x = 1\)) as\( h \to 0\)
Limit
Calculas
trigonometry
Answer: does not exist
TOMATO, Problem 728
Challenges and Thrills in Pre College Mathematics
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 ..........
Now, lim \(2tan^-(1/h)\) as\( h \to 0\) doesn‟t exist
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