Calculus and the world.

Level 2

If f ( X ) f (X) = X^2 + { 1+(tan 45˚)^2) }, and when X=2, f ( X ) f (X) = 2.5. So, what limit is X approaching when X=2, and what is f ( X ) f (X) equal to when X equals to 1.795?

2.5, 5.434 6, 5.354 6, 5.222 2.5, 4.22

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1 solution

Tejas Chakrabarti
Jul 27, 2018

So firstly, f ( X ) f (X) = X^2 + {1+ (tan 45˚)^2}, We know that tan 45= 1, and when {1+ (tan 45˚)^2} is the function, we can merely reduce it to (sec 45˚)^2, which is equal to 2. So, whatever may be the value for X, we know that f ( X ) f (X) is equal to X^2+2. Finally the limit to the function f ( X ) f (X) when X is equal to 2 is not 2.5, but rather 6, as although when one plots a graph, and upon reaching two on the X axis, the value will be 2.5, but the limit to it does not regard the special rule applied to it. Also, the square of 1.795 is 3.222, and when this number is added to 2, the answer is 5.222.

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