\[f\left( x \right) f\left( y \right) =f\left( x \right) +f\left( y \right) +f\left( xy \right) -2\]

If \(f\left( x \right) \) is a polynomial satisfying the equation above for all \(x\) and \(y\) and \(f\left( 2 \right)=5 \), then compute \(\displaystyle \lim _{ x\rightarrow 2 }{ f^{ ' }\left( x \right) } \).

If anyone solves this one, please do tell me what is the error in this one

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## Comments

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TopNewestHey there!

We have,

\(

f(x)f(y)-f(x)-f(y)+1\quad =\quad f(xy)-1\\ (f(x)-1)(f(y)-1)\quad =\quad f(xy)-1\quad \\ g(x)=f(x)-1 \\ \Rightarrow g(x)g(y) = g(xy) \\ \)

This is famous, see here.

We then have that,

\( g(x) = x ^ t, t = 2. \\ g(x) = x^2 \\ f(x) = x^2 + 1 \\ f'(x) = 2x \\ \Rightarrow \lim _{ x\rightarrow 2 }{ f^{ ' }(x) } = 4 \)

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Put x and y as 2 which will give you the value of f(4) as 17

Put x as 4 and y as 2 and you get f(8) as 65. Similarly you can check with other values and this leads to the conclusion that f(x) is one greater than the square of x.

Hence the limit computation will be 4

Please excuse me if you were looking for a formal proof or evaluation of the same.

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@parv mor

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@Rishabh Cool, this question again.

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