How do you evaluate #(e^(4x) -1 -4x) / (x^2)# as x approaches infinity?

1 Answer
Apr 23, 2016

The answer is infinity.

Explanation:

It takes some logic to solve this one. Plug in infinity into the equation...you'll get:

#(e^(4∞)-1-4(∞))/(∞)^2#

Infinity squared? That's a huge number. But something to the power of infinity? That's even bigger.

Remember, infinity is simply a place holder for a large number...I perceive the above equation as such...

Super mega large number / large number = #∞/1#

The answer comes out as #∞#. Graph and see for yourself.