Question

limits

Answer 1

\(\lim\limits_{v \to \infty} f(v) = v^2 - 2v(1+ \sinh v)\cosh v + 1 + 2 \sinh v\)

best i can muster is the obvious expansion ....

\(= \lim\limits_{v \to \infty} v^2 - 2v(1+ \dfrac{e^v - e^{-v}}{2})(\dfrac{e^v + e^{-v}}{2}) + 1 + (e^v - e^{-v}) \)

a deletion...

\(= \lim\limits_{v \to \infty} v^2 - 2v(1+ \dfrac{e^v}{2})(\dfrac{e^v}{2}) + 1 + e^v \)

...and the observation that the dominant term is \(- v e^{2v}\) and so the limit is \(-\infty\).

works for me, but not very mathsy.

is there a better/proper way to do this?

Answer 2

thanks. that looks way too good, i'm sticking with my hand waving :-)

Answer 3

i would stick to hand waving too, wolfram is being too critical haha