limit tends to x---3 [1/x-3] - [3/ x (x^2 - 5x + 6)]

here x-3 is fully in denominator and x (x^2 - 5x + 6) is also in denominator

Question

limit tends to x---3 [1/x-3] - [3/ x (x^2 - 5x + 6)]
here x-3 is fully in denominator and x (x^2 - 5x + 6) is also in
denominator

Pranesh Kumar · Accepted Answer

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limit tends to x---3 [1/x-3] - [3/ x (x^2 - 5x + 6)] here x-3 is fully in denominator and x (x^2 - 5x + 6) is also in denominator

limit tends to x---3 [1/x-3] - [3/ x (x^2 - 5x + 6)]

here x-3 is fully in denominator and x (x^2 - 5x + 6) is also in denominator