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in Continuity and Differentiability by (27.0k points)
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If f(x) = \(\frac{tan(\frac{\pi}{4}-x)}{cot\,2x}\) for x ≠ \(\frac{\pi}{4}\), find the value which can be assigned to f(x) at x = π/4 so that the function f(x) becomes continuous every where in [0, π/2].

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A real function f is said to be continuous at x = c, where c is any point in the domain of f if :

where h is a very small ‘+ve’ no.

i.e. left hand limit as x → c (LHL) = right hand limit as x → c (RHL) = value of function at x = c.

This is very precise,

We can summarise it as, 

A function is continuous at x = c if  :

Function is defined for [0,π] and we need to find the value of f(x) so that it is continuous everywhere in its domain 

(domain = set of numbers for which f is defined)

As we have expression for x ≠ π/4, 

which is continuous everywhere in [0,π],

so If we make it continuous at x = π/4 

it is continuous everywhere in its domain. 

Given,

 f(x) = \(\frac{tan(\frac{\pi}{4}-x)}{cot\,2x}\) for x ≠ \(\frac{\pi}{4}\) …….equation 1

Let f(x) is continuous for x = π/4

[multiplying and dividing by π/4–x and π/2–2x to apply sandwich theorem]

∴ value that can be assigned to f(x) at x = π/4 is \(\frac{1}{2}\)

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