how do I find the limit of a function?
608336
L'hopital's Rule does not apply in this case as the limit can be evaluated as it is, it is not one of the special cases which you would apply that (infinity / infinity, zero / zero).  Actually, if you did apply L'hopital's Rule, you would get the wrong answer (wrong answer being 1).\n\nIn this case, since as x approaches 3 (from either side), the numerator of the function goes to zero and the denominator goes to some finite, no zero quantity, you can just plug in 3 into the function and evaluate.\n\nThe limit of the function as x --> 3 = 0
1047841
By the def\n\nlim f(x + h ) - f(x)\n              /h\nPlug in (x+h) into f\nlim (x+h)^2 + 5(x +h) - (x^2 + 5x)\n                        /h\nFoil\nlim x^2 + 2xh + h^2 + 5x + 5h - x^2 - 5x\n                          /h\n\nCancel like terms\nlim 2xh + h^2 + 5h\n           /h\n\n\nReduce everything by h\nlim 2x + h + 5\n\nTake the limit and\nPlug in 0 for h\n2x + 5
784830
just another math guy may be correct, but there is a much simpler solution using l'Hospital's rule.\n(In this proof, all limits are as n→∞)\nf(n) = (1+1/n)^n = [(n+1)/n] ^ n\n= (n+1)^n / n^n\nTake the derivitive of numerator and denomerator to find an equivalent limit.\nlim f(n) = lim [(n+1)^n * (n/(n+1) + ln(n+1))] / [n^n * (1+ln(n))]\n= lim (n+1)^n / n^n * { [n/(n+1) + ln(n+1)] / [1+ln(n) ] }\n= lim f(n) * { [n/(n+1) + ln(n+1)] / [1+ln(n) ] }\n\nTherefore, lim { [n/(n+1) + ln(n+1)] / [1+ln(n) ] } = 1 !\nor lim n/(n+1) + ln(n+1) = lim (1 + ln(n) )\nor, in the limit, \nln(n+1) - ln(n) = 1/(n+1)\nln[(n+1)/n] = 1/(n+1)\ne^(1/(n+1)) = (n+1)/n = 1+1/n\nor e = lim (1+1/n)^(n+1) = lim (1+1/n)^n
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.
742785
as x ->pi/4 (1-tan x) ->0 and sec x is not infinite(sqrt(2))\n\nso limit is product =0
953356
Let f(x) = sin (1/x) and g(x)=-sin (1/x). Neither f nor g have limits at 0, but [x→0]lim f(x) + g(x) is obviously 0.
798431
limit x>0 [(3x)sin3x/3x*(4x)sin4x/4x]\n/[(6x)tan6x/6x(x)sinx/x]\nlimit x>0 3x*4x/6x*x\nlimit x>0 12x^2/6x^2\n=2 (answer)
892605
(constant)/(2^n) when n->infinity => the limit is 0(zero)\n\nPS\nNot to be a a stickler for rules but is this really laptop related ? :))
692865
V = x * (4-x) * (6-x) = x^3 - 10x^2 + 24x\n\n0 < x < 4\n\nGraph the function and find the max between 0 and 4
477866
Let y=cos2x, then, \ndy/dx = limit (as h tends to zero) [(cos2(x+h) - cos2x)/h]\n      = lim. [(cos2x cos2h  - sin2x sin2h - cos2x)/h]\n      = lim. [(cos2x(cos2h-1) -sin2x sin2h)/h]\n      = lim. [(cos2x (-2(sin2h)^2)/h] - sin2x* lim.(sin2h/h)\n      = -2 cos2x * (lim. sin2h/h)*(lim. (sin2h))- 2 sin2x\n      = -2 cos2x * 2 * 0 - 2 sin2x\n      = -2 sin2x
778101
x^x = e^{xln(x)} \nnow, the lim x->0 x ln(x) = 0,\nso lim x->0 x^x = e^0 =1\n`\n`
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.
491386
Just substitute x+1 into f(x)\n\nf(x+1) = [ 6(x+1)^3 ] - [ 3(x+1)^2 ] + 4(x+1) - 9
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
723571
A function f(x) is continuous at a point (x=a), only if:\n\nIt's Left-hand side limit equals it's Right-hand side limit.\n\nDefinition:\n\nLHS limit at (x=a):\n\nlim(h->0)  f(a-h)\n\nRHS limit at (x=a):\n\nlim(h->0)  f(a+h)\n\nOnly if these two limits are equal to each other, then the function f(x) is continuous at the given point (x=a)\n\n\n\n\nLet me give you an easy example:\n\nSuppose our function is f(x)=(1 / x)\n\nWe want to check whether this function is continuous at (x=0).\n\nSo, first we start by taking it's LHS limit:\n\nlim (h->0)  f(a-h)\n= lim (h->0)  f(0-h)\n= lim (h->0)  (1 / (0-h))\n= lim (h->0)  (-1/ h )\n\nNow, keep this as it is & proceed to find the RHS limit:\n\nlim (h->0)  f(a+h)\n= lim (h->0)  f(0+h)\n= lim (h->0)  (1 / (0+h))\n= lim (h->0)  (1/ h )\n\nNow, we compare our LHS and RHS limits, we find that both of them are almost the same - BUT, the LHS limit has a negative value & the RHS limit has a positive value. Hence, they are not completely equal.\n\nTherefore, we say that our function f(x) = (1 / x) is NOT CONTINUOUS at (x=0).
888635
Let f(x) = 3x^4 - 25 + 5/4. Consider an arbitrary real valued x say x = c. We need to determine if the left-hand behavior and the right-hand behavior of the function f at x = c are the same or not. \n\nBecause f(x) is a polynomial function (of degree 4) on x, it follows that the function f is continuous on R (the set of real numbers).  So, if we take the limit on the left (and right) of x, we will get the same value for both sides. \n\nThat is, the behavior on the left (Left-hand Limit) will be: as x approaches  c from the left, we get f(x) approaches f(c) = 3c^4 - 25 + 5/4. Similarly, the behavior on the right (Right-hand Limit) will be: as x approaches c from the right, we also get f(x) approaches f(c) = 3c^4 - 25 + 5/4. \n\nTherefore, both the left and right-hand limit yields to f(c), i.e. as x approaches c, we always obtain f(c).\n\nHence, both left and right-hand behavior of f are the same for all real valued x.