Suppose f is convex and differentiable on R. If lim as x --> - infinity of f(x) = - infinity,?

Question

then show that lim x--> infinity of f(x) = infinity.

Please show your work. thank you for your help

Answer 1

The statement that f is convex and differentiable on R implies that f' (the derivative of f) is monotonically non-decreasing on R.
It also implies that f is continuous on R.

lim[x -> -infinity] f(x) = -infinity means that for any z, there exists an x such that x < z and f(x) < f(z).

Choose any value for z. Choose a corresponding x, such that x < z and f(x) < f(z)

The mean value theorem tells us that there exists a y, such that:
f'(y) = ( f(z) - f(x) ) / (z - x)

Therefore, f'(y) is a positive number. f'(y) > 0

f' is monotonically non-decreasing. That implies that for all z such that z > y, we have f'(z) > 0

Thus...
lim[x -> +infinity] f(x) = integral[y to +infinity] f'(z) dz + f(y)
and f'(z) is always positive on the interval y to +infinity.

This implies that
lim[x -> +infinity] f(x) = +infinity

Answer 2

lim t ? ? [ ? sin(x) dx from t to ?t ] = lim t ? ? [ ? cos(x) | x=t to ?t ] = lim t ? ? [ ? cos(t) + cos(?t) ] = lim t ? ? [ ? cos(t) + cos(t) ] = lim t ? ? [ 0 ] = 0 i won't be in a position to work out how the imperative of sin(x) dx over ?t to t, with t ? ? is undefined. pondering sin(x) it fairly is an choppy function, comparing that is imperative over any symmetric era will continually yield 0. however the seize lays in this final condition, that it fairly is a symmetric era. An imperative from ? to ?? isn't comparable to an imperative from t to ?t with t ? ?. you won't be in a position to assume that the two infinities are approached at equivalent "fees". hence, for the 2d question, you may desire to chop up the imperative: ? sin(x) dx from ? to ?? = ? sin(x) dx from ? to 0 + ? sin(x) dx from 0 to ?? = lim r ? ? [ ? sin(x) dx from r to 0 ] + lim s ? ? [ ? sin(x) dx from 0 to ?s ] = lim r ? ? [ ? cos(x) | x=r to 0 ] + lim s ? ? [ ? cos(x) | x=0 to ?s ] = lim r ? ? [ a million ? cos(r) ] + lim s ? ? [ cos(?s) ? a million ] = lim r ? ? [ a million ? cos(r) ] + lim s ? ? [ cos(s) ? a million ] = lim r ? ? [ ? cos(r) ] + a million + lim s ? ? [ cos(s) ] ? a million = lim r ? ? [ ? cos(r) ] + lim s ? ? [ cos(s) ] the two words do no longer converge (as r ? ?, the fee of the 1st term will save oscillating between ? a million and a million and in addition for the 2d term as s ? ? between ?a million and a million). For any pair of r and s as they attitude ?, the imperative would have a cost between ?2 and a pair of. consequently this imperative isn't defined. the apparent contradiction stems from the reality that for the time of the 1st case, the area of integration exchange into symmetric, as the two ends approached ? and ?? the two (watching an identical variable, t). in the 2d case, this assumption would desire to no longer be made, and the fee of the imperative exchange into undefined.