The value of this as it tends to infinity ((-e-a + sqrt((e-a)^2 + 4 x pi c ln(61.3)))/2)

Question

I hope this is the right place to post...

I'm not strong on my 'infinity maths' and in doing some challenge eigen vector question a freind set me I ended up with this:

(-e-a + sqrt((e-a)^2 + 4 x pi c ln(61.3)))/2 as a tends to infinity

Anyone with any ideas on what the value of this is as I'm stumped. I'm expecting it to be infinity or -infinity I think but there's -infinities and +sqrt(infinities) so I got no idea

Answer 1

>4 x pi c ln(61.3))

These are all constants, so just abbreviate it with `C` or something. Then you just have

        -e-a + √[(e-a)² + C]
    lim --------------------
    a→∞          2

You can evaluate this by **multiplying with the conjugate** of the numerator

        -e-a + √[(e-a)² + C]   -e-a - √[(e-a)² + C]
    lim -------------------- · --------------------
    a→∞          2             -e-a - √[(e-a)² + C]

Continue from here for now

Answer 2

> I'm expecting it to be infinity or -infinity I think but there's -infinities and +sqrt(infinities) so I got no idea

That will often be the case, but not here. You have 'a' in two places: -a, and sqrt(a^(2) and other stuff): the sqrt(a^(2) and other stuff) is going to approach a, and the -a and a will almost cancel out.

In general, if you have something that looks like -x + sqrt(x^2 + bx + c), you complete the square:

    -x + sqrt(x^2 + bx + c)
    -x + sqrt( x^2 + bx + b^2/4 - b^2/4 + c)
    -x + sqrt((x + b/2)^2 + c-b^2/4)

Now the square root term approaches x+b/2 as x-->infinity, the -x and +x cancel, and the whole thing approaches b/2.

You have essentially that same question here: your c, e, x, 2, pi, 4, and 61.3 are all constants.