How to prove this function is unbounded as x goes to plus/minus infinity?
146506
well, you don't have to worry about sinx and cosx as they are always between -1 and +1. so the first part of the function (1/2)* sinx is bounded.\ncos(x) is bounded.\nSo, you can rewrite the function as,\ny(x) = f(x) - x*g(x).\nwhere both f(x) and g(x) are bounded.\nso, as x goes to infinity, y(x) tends to minus infinity because of -x in the right hand side. \nSimilarly when x goes to minus infinity, y(x) goes to plus infinity\n----------------------\nResponse to the counter-example - \nin case y(x) = 2+x*(-3),\nas x -> infinity y -> -infinity.\nso y(x) is unbounded\nwhat's wrong with that??????
615960
As the graph for y=logx will approach but never reach x=0, the value for log0 is undefined. However, it does APPROACH negative infinity.
819082
Let f(x) = x ³/ (x² + 2x - 15) \n\n= x - 2 + (19x -30)/(x² + 2x - 15) (by division)\n\n= x - 2 + (19x -30)/[(x - 3)(x + 5)]\n\nSo f(x) is undefined when x = -5, 3\n\nAlso lim x→±∞ f(x) = x - 2\n\nThus f(x) is asymptotic to g(x) = x - 2 as x →±∞ and asymptotic to x = -5 and x = 3
904469
Let u = e^x + 1.  Then du= ...\n\nHope that helps.
952602
This is what always helps me.\n\nA function is a machine.\n\nThe domain is everything you can put into the machine so that it doesn't break.\n\nThe range is everything you can get out of the machine.\n\nIn the case of your function, there's only one possible value of x that will "break" the function, in the sense that you'd get an infinite answer.  That value is x=0, because then you'd have 9/0 + 7, and 9/0 is infinity.\n\nSo the range is {x | x is not equal to 0}.\n\nAssuming that you meant 9/x + 7, rather than 9/(x+7).  If you meant the latter, x cannot be equal to -7.
857286
Im not sure don't right this at your Homework.........\nwhere f(x) > 0. a. (–3, 1) b . (–3, 0) ∪ (1, ∞) c . (1, 3) d . (0, 1) ∪ (3, ∞) If f(x) = x(x – 1)(x – 4)2, use interval ..............??????\n\n\nSORRY I DON'T HELP TOO MUCH I TRY MY BEST......
62352
t can be any from negative infinity to positive infinity.  It doesn't matter because the 0 in the numerator makes this equation zero, ALWAYS.  The only thing t cannot be is 0, since the denominator cannot be zero.  That would make the equation "undefined."\nWas that your question?
490253
infinity/0 is infinity.  If you have any function f(x) and g(x) where f(x) goes to infinity as x -> k and g(x) goes to zero as x -> k, then 1/g(x) goes to infinity as x -> k, so\n\nf(x)*(1/g(x)) -> infinity as x -> k.\n\ninfinity*zero depends. If you look at 1/x and x^2, then as x -> 0, 1/x -> infinity and x^2 -> 0, but\n\n(1/x)*(x^2) = x, so its limit is 0 as x -> 0.\n\nIf you look at 1/x^2 and x, then as x -> 0, 1/x^2 -> infinity and x -> 0, but\n\n(1/x^2)*(x) = 1/x, so its limit is infinity as x -> 0.
579155
The first one isn't, because it's undefined when x=4 (assuming the (x-4) is the denominator, not just the x)\n\nThe second one is; it's a horizontal line with no breaks or holes.
804449
There is no limit. The problem is that sin(x) goes between postive one and negative one infinitely many times as x gets small (i.e. goes to negative infinity). The hard part, though, is to figure out how to show this.\n\nI will look at two different sequences of numbers that go to negative infinity. When I plug the numbers in the first sequence into x/(2sin(x)) they will go to positive infinity. When I plug the numbers in the second sequence into x/(ssin(x)) they will go to negative infinity. If I can get two different answers then there is no limit.\n\nThe first sequence is x_n = -pi/2, -pi/2-2pi, -pi/2-4pi, -pi/2-6pi, ....,-pi/2-2n pi\n\nWhen I plug this into x/(2sin(x)) I get\n(-pi/2-2n pi)/(2sin(-pi/2-2n pi))=(-pi/2-2n pi)/(-2)=pi/2+2n pi\nAs n gets big then this number goes to positive infinity.\n\nThe second sequence is x_n = -3pi/2, -3pi/2-2pi, -3pi/2-4pi, -3pi/2-6pi, ....,-3pi/2-2n pi\n\nWhen I plug this into x/(2sin(x)) I get\n(-3pi/2-2n pi)/(2*sin(-3pi/2-2n pi))=(-3pi/2-2n pi)/(2)=-3pi/2-2n pi\nAs n gets big then this number goes to negative infinity.\n\nWe now have two sequences of numbers that give different answers. The function cannot converge to a single number.
918808
wouldnt it be like 3<x<-3 f(x)\nso like x would be 1 so it would be just f if you really want to simply that far done.
1154919
<1>...u know lim xsin(1/x)=lim (sin(1/x))/(1/x),and when x goes to infinity ,(1/x) goes to <0>....and we know that (sina)/a when <a> goes to <0> is 1...and here <a> is <1/x>...so it's limit is <1>
208997
The only way to get the expression infinity/-infinity is through a limiting process in the real numbers, so I'll assume you're talking about limits. In that case the form above is indeterminate, meaning that it depends on the functional forms in the limit. \nSuppose you have f(x) and g(x) such that lim [f(x)/g(x)] = infty/-infty. Consider the equivalent expression lim [f(x)/g(x)] = lim [(1/g(x))/(1/f(x))] which approaches the form 0/0. This allows the use of L'Hospital's rule to evaluate the limit, which we know from experience can be anything.
566883
First, since 1/x is not defined for x = 0, before you can show  anything about your functions, you have to define values for g(0) and f(0). I'm going to assume that either the original statement of the problem had that or that you are supposed to find the value that makes the problem solvable.\n\nTo be true you have to define the problems with f(x) = 0 and g(x) = 0\n\n\nFor the continutiy of f(x) given epsilon > 0 find delta > 0 such that |f(x) - f(0)| < epsilon for all |x-0| < delta.\n\n|f(x) - f(0)| = |xsin(1/x) - 0| < |x| = |x - 0|\n\nGiven epsilon, pick delta = epsilon.\nIf |x-0| < delta, then |f(x) - f(0)| < epsilon.\n\n\nAs for the differentiability of g(x),\n\nf(0+h) - f(0)) = (0+h)^2sin(1/(0+h)) - 0\n\nlim(x->0)(f(0+h) - f(0))/h = \nlim(x->0)((0+h)^2sin(1/(0+h)) - 0)/h = \nlim(x->0) h^2(sin(1/h)/h = lim hsin(1/h) = 0\n\nSo the limit exists and is 0.
708189
the integral diverges,\nsince lim x->+-infinity e^(pix^2)  = infinity\ns
845874
1.  This is equal to\nk*(lnx)/x, and since \n(lnx)/x --> 0, so does this expression.\n\n2. Sorry, blank on this and 3.\n\n4. (n^k)/2n = (n^(k-1))/2, and I guess as long as \nk > or = 1\nthis will approach 0.\n\nOf course, if k < 1 then it has no limit, or "approaches infinity"
653191
You have to look for the vertical asymptopes. For exapmle in (x^2-5)/((x+2)(x-5)), the vertical asymptopes are the ones that will make the denominator 0, so -2 and 5. With these, as the function approaches them, it will rise to infinity, or unless you have a horizontal or slant asymptope.