finding the limit using l'Hospital's Rule if necessary?
921908
1\n\njust do a little bit of trial you will see that the smaller number you use it nears 1.
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"
748556
L'Hospita;'s rule shall not apply here as it is non zero/zero form\n\nit applies for form 0/0 or infinite.infinite\n\nlimit f(x)/g(x) when it is of above form is lim f'(x)/g'(x) \n\nfor example as x->o lim sin x/ x is of the form 0/0\n= lim x->0 d/dx(sin x)/d/dx(x)\n= cos x/ 1= cos x = 1 when x->0\n\nfor your case it is 1/0 or infinite (no l'Hospital's rule)
98350
limit as x tends to 3 ;( x^2 -3)/2 =(3^2 -3)/2 =(9-3)/2 = 3
334668
Nasty one.  Treat both parts of the sum separately.\n\nPart 1 lim(x to 0) (sin x)^(1/x):\n\nLet y=(sin x)^(1/x). Take log (base e) of both sides so\n\nln y=(1/x) ln(sin x)\n\nRemember that log of the limit of something is the same as the limit of the log.  So ln(lim(x to 0)y)=lim(x to 0)(ln y)=lim(x to 0)(ln(sin x)/x) which does not exist (but heads off at the minus infinity end).  Hence, taking exponential of both sides we get lim(x to 0)y=e^(-infinity)=0.  So the left hand term gives us zero at the limit.\n\nPart 2 lim(x to 0) (1/x)^(sin x):\nThis one is harder and if you haven't heard of L'Hopital's rule then you may be in trouble (although there might be a way to do it anyway).\n\nUse the same idea of taking log base e of both sides of y=(1/x)^(sin x).  So\n\nln y=sin x ln(1/x).\n\nSo, lim(x to zero)ln y=lim(x to zero)(ln(1/x))/(1/sin x).  x=0 yields infinity over infinity so to speak for that quotient so we can use L'Hopitals rule and differentiate top and bottom with respect to x to get\n\nlim(x to 0)ln y=lim(x to 0)(-1/x)/(-cos x/(sin x)^2)=lim(x to 0)((sin x)^2)/(x cos x)\n\nWe now get 0/0 so we use L'Hopital one last time to get\n\nlim(x to 0)ln y=lim(x to 0)(2sin x cos x)/(cos x-x sin x) which equals zero\n\nSo ln(lim(x to 0)y)=lim(x to 0)(ln y)=0.  Taking exp of both sides gives\n\nlim(x to 0)=e^0=1\n\nNow add both limit terms together (0+1) and you get your final answer\n\n{(sinx)^1/x} + (1/x)^sinx=1\n\nHope that was clear enough to follow.\n\nLuke
737185
How about something simpler, like\n\nf(x) = 0 when x<1/2, 1 when x>=1/2\n\nThen the limit from the left is 0, the limit from the right is 1, so the regular limit doesn't exist.
945609
Use l'Hospital's Rule when dealing with 0/0 or intfy/infty: in such cases,\n\nlim (f(x)/g(x)) = lim (f'(x)/g'(x)) if the latter exists (you may have to repeat this step using the second derivatives f''/g'').\n\nIn your first example, as x --> 1, \n\nlim (x+ x^2 + x^3 +... + x^n - n)/(x-1) \n= lim (1+ 2x + 3x^2 +... + nx^(n-1) )/1 \n= 1+ 2 + 3 +... + n = n(n+1)/2.
1114168
Ok, here's how i think you do it.\n\nto get the horizontal asymptote, you have to get the limit of the functon when x tends to be infinite, so you would get something like this\n\nlim     (3x-x^2)/(4-x)\nx→∞\n\nthen you substitute ∞ on the x's and you get\n\n∞/∞, that's not defined, so what you use is l'hopital law where you derive both the nominator and the denominator of a fraction, so you will get\n\nlim    3-2x/-1\nx→∞\n\nnow you can substitute ∞ and you get\n\n(3-2*∞)/-1 =∞\n\nand you do the same with -∞ and you get as an answer -∞\n\nwhich means there is no horizontal asymptote.\n\nI hope this helps.
1132233
this question will require calcs using logarithm.\nhence will take time
691996
Let u = lnx, then du = (1/x)dx\nln(e) = 1 and ln(infinity) = infinity\nint from 1 to infinity of 1/(u*sqrt(u)) du\nu*sqrt(u) = u*u^(1/2) = u^(3/2)\nint from 1 to infinity of 1/u^(3/2) du\n= int from 1 to infinity of u^(-3/2) du\n= limit as b goes to infinity of -2u^(-1/2) from 1 to b\n= limit as b goes to infinity of -2(b^(-1/2)-1)\n= -2(0-1)\n= 2
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
665272
To build on what "Thin Kaboudit" was saying, you can take the limit of a function f(x) = 3/x as x goes to zero.\n\nLets say you decided to approach x from the positive side, slowly decreasing its value toward zero.\nThe closer and closer x gets to zero, the closer the function's value gets to [positive] infinity.\nNow lets say you decide to approach x from the negative direction, starting out with negative numbers and slow increasing them toward zero.  As you get closer and closet to zero from the negagtive side, the function goes toward negative infinity.\n\nSince the limit from the positive side does not equal the limit from the negative side (infinity does not equal negative infinity), the limit of the function does not exist since the function is not defined when x = 0.
689918
use  power rule and product rule\nf'(x)=[1/5]*5*x^4 + 2x(x^2-x-9)+ \n(x^2+9)(2x-1) + 0=\n\nx^4 + 2x^3 - 2x^2 -18x + 2x^3 - x^2 +18x -9 =\n\nx^4 + 4x^3 - 3x^2 -9
221164
to get an extra credit, maybe?\n\n---\n\nseriously, I cannot figure out how you can study calculus without knowing limits: even if you only deal with numerical procedures, every iterative formula, or better, EVERY FORMULA (such as those used to find the roots of a non-linear equation, approximated integrals, derivatives and so on) needs a preventive study for convergence conditions: this involves limits!\n\nOne more step back: DERIVATIVE IS A LIMIT ! How can you apply it without knowing what it is?
788373
A removable gap can occur when the limit to a point outside the function domain exists. 1 is outside the domain since the denominator is 0 for x = 1. In this case, what you need to do is show that \n\nlim[x->1] f(x) exists. \n\nf(x) = ln(x^2)/(x^2-1) = 2ln(x)/((x + 1)(x - 1))\n\n= 2/(x + 1)*(ln(x)/(x - 1))\n\nIt is this second factor we care about, Using L'Hospital's law, top -> 0, bottom -> 0, so the limit of ln(x)/(x - 1) = lim (1/x)/1 = 1, so the limit of the expression is 1.
788532
Is it x*(e-2) or x^(e-2)?\n\nI don't think it matters anyway, because the bottom fluctuates between 3 and 5 while the top increases indefinitely, i.e. approaches "infinity" if you like.  So there's no limit.
1016907
As you specified that limit as 'x' tends to (pi/4), I have changed the solution accordingly.\n\nLim  x--> (pi/4) [{sin(x) - cos(x)}/cos(2x)]\n= Lim  x--> (pi/4) [{sin(x) - cos(x)}/{cos^2(x) - sin^2(x)}]\n= Lim  x--> (pi/4) [{sin(x) - cos(x)}/{(cos(x) + sin(x))*(cos(x) - sin(x))}]\n= Lim  x--> (pi/4) [-1/{cos(x) + sin(x)}]\n=  [-1/{cos(pi/4) + sin(pi/4)}]\n=  -1/[{1/sqrt(2)} + {1/sqrt(2)}]\n= -1/[2/sqrt(2)]\n= -{sqrt(2)}/2 \n\nYou don't need to use L.Hospital's rule.