Calculus Limit Problem?
540790
You have two cases.\nCase 1:\nIf c is negative, then e^(cx) goes to 0 as x goes to infinity.  The limit would then be zero.\nIf c is positive, then you can apply L'Hopital's rule.\nIt becomes\n[bce^(cx)/(a+be^(cx))]/d = bce^(cx)/[d(a+be^(cx))]\nThis also goes to infinity over infinity.  Use the rule again.\n[b(c^2)e^(cx)]/[dbce^(cx)]\nYou can cancel b, c and e^(cx) to get\nc/d\nAnswer:\nIf c < 0 , then the limit is 0.\nIf c > 0, then the limit is c/d.
203181
I'll interpret this as f(x,y)=x^3 y/(x^2 + y^2). I would rewrite it as f(x,y)=xy [x^2/(x^2 +y^2)]. As (x,y)->(0,0), the first term goes to 0 (clearly), while the second term is bounded (by 1), so the limit is 0.
632327
Write f(x,y) as x^2*[y^2/(x^2+y^2)]. The second factor is always <=1, so f(x,y) goes to 0 as (x,y) goes to (0,0). The first answerer is WRONG. As for the second answerer, just knowing the limit is 0 along every line is not enough to conclude that the limit exists. It is possible for limits along lines to be 0, but to be otherwise along parabolas.\n\nIf f(0,0) is defined to be 0, then this function has partial derivatives at (0,0) that are both 0. In fact, f is differentiable at (0,0) in that case. The derivative is the 0 functional.
725168
Use L'Hospitals rule\nd(sin ax)/d( bx)\na*cos(ax)/b lim x----> 0  = a/b
954661
Suppose f : D → R has a limit at x0. Prove that | f | : D → R has a limit at x0 and that limx→x0 | f(x) | = | limx→x0 f(x) |.\n\n\nLet limx→x0 f(x) = c\n\nNow | f(x) | = f(x) for f(x) ≥ 0 = c\n= -f(x) for f(x) < 0 = -c\n\nSo limx→x0 | f(x) |  \n\n= limx→x0 f(x) for f(x) ≥ 0 = c\nlimx→x0 -f(x) for f(x) <0 = - limx→x0 f(x) x < 0 = -c\n\n= |c|\n\n= | limx→x0 f(x) | .................... QED
475495
JohnyBoy almost has it right.  When you multiply both numerator and denominator by sqrt(1+cosX), the numerator becomes    X sqrt(1+cosX)        , not X (1+cosX),\nThe rest is correct.  So the limit is sqrt(2).
274355
Depends on the system you are trying to solve. check your limits of integration. from -infinity to 0 plus 0 to infinity.
487867
Here is the mathematical proof:\n\nIt involves some precalculus, so I will explain what I am doing just in case you haven't gotten to precalc yet.\n\n[1]  Definition: Limit\n     The limit of a function f(x) is the value of y @ x.\n     i.e. y = 2x^2, @ x = 1, y = 2, \n     Therefore, \n     limit 2x^2 = 2\n     x->1\n\n[2]  Why division by Zero = infinity:\n\n     limit 1/x = infinity\n     x -> +0\n\n     Limit 1/x = infinity\n     x -> -0\n\nThe graph of y = 1/x explains this principle. \n\n[2.1] Definition: +0  & -0\n\n+0 = As x approaches zero from the right.\n-0 = As x approaches zero from the left.\n\n[2.2] As x get closer and closer to zero, the quotient of 1/x gets larger and larger.\n\ni.e. 1/0.0000006 = 6000000\ni.e  1/1x10^-9999= 1x10^9999\n\n[2.3] When x becomes zero, the quotient of 1/x is infinity. Since infinity is underfined, 1/0 is also undefined!
787424
To find the asymptote to the left.  find the limit of f(x) as x approaches -∞.\n\nTo find the asymptote to the right.  find the limit of f(x) as x approaches ∞.\n\nVertical asymptotes usually occur where a value of x gives a division by zero.  Find when the denominator equals zero then take the limit as x approches that value to make sure the value of the function approaches negative or positive infinity.\n\n I hope that helps.
1023164
Limit ( n to infinity) (n + i/n)^ n
921398
I just taught this section today to my calc students.  You start by finding the first and second derivative.  I recommend looking in your book for an example.  Your book probably has something like this towards the front of the section.  It would be too long winded to write everything down here.  If you don't have your book on you, the web page below has problems that you can do and then check your answer.\n\nhttp://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/graphingdirectory/Graphing.html
613021
http://www.ask.com/web?q=limit%28%28%28n%21%29%5E%281%2Fn%29%29%2Fn%29%2Cn%3Dinfini...+&qsrc=0&o=0
377994
Is this (x + 3)/[x*(x - 1)^2] ? In that case, the limit is as you stated, positive infinity. \nA limit is never undefined; it either exists, or it does not exist. In the special case that it does not exist and it increases without bound or decreases without bound from all approaches, we say the limit is positive or negative infinity respectively.\nAs always, when a limit diverges, look at the behavior of the expression in an arbitrarily small region around the limit point (in this case, x=1). Since the denominator is a square multiplied by x, it is never negative in a sufficiently small neighborhood of 1 (specifically, any neighborhood contained in the interval (0,infinity) ). Similarly, the numerator is never negative in a sufficiently small neighborhood of 1, so the behavior of the whole expression is to diverge to positive infinity, no matter how it is approached.
914503
You should remember how to do this in the future because the limit of this expression as h approaches 0 gives you the derivative or slope of f(x) at x = 4.\n\n[ f(4 + h) - f(4) ] / [ (4 + h) - 4 ]\n\n[ (4 + h)^2 - 4^2 ] / h\n\n(h^2 + 8h + 16 - 16) / h\n\nh + 8
891612
You can convert f(x) as follows:\n\nf(x) = 1 / SQRT(4*(1+x/4)) = 1/[SQRT(4) * SQRT(1+x/4)]\nf(x) = 0.5 * 1/(1+x/4)^(1/2) = 0.5 * (1+x/4)^(-1/2)\n\nUsing the approximation:\nf(x) ~ 0.5 * (1 - x/8) = 0.5 (when x--> 0)\n\nThe only thing I don't understand is why you need the approximation.  The limit can be calculated straightforward without it.
454694
1\nDetails:\nLet y = lim x->0 of (sinx)^x\nTake the natural log of each side.\nln(y) = lim xln(sinx)\nWrite xln(sinx) as (ln(sinx))/(1/x)\nas x goes to 0, this goes to infinity over infinity.\nUse L'Hopital's rule \n(ln(sinx))/(1/x) becomes\n(cosx/sinx)/(-x^(-2))\n= (-x^2)(cosx)/(sinx)\nAs x goes to 0, this goes to 0/0.\nUse L'Hopital's rule  again\n(-x^2)(cosx)/(sinx) becomes\n[(x^2)(sinx) - 2x(cosx)]/(cosx)\nPlug in 0.\n[0-0]/1 = 0\nThis equals ln(y)\nln(y) = 0 means y = 1
962773
Use L'Hospital's rule twice and get e^x/2 which has a limit of infinity.