how to get the answer of lim (x-5y)/(x^2 + y^2) when x and y tends to 0?
476547
First, make the substitution y = mx.  That will give you\n\n[x - 5mx]/[x^2 + (mx)^2].\n\nNow, factor out the x term and you get\n\nx[1 - m]/(x^2)[1 + m^2] = [1 - m]/x[1 + m^2]\n\nAs you can see from this expression, changing the value of m will change the value of the limit.  For a limit to exist, it must be unique.  Therefore, since the limit depends upon the value of m, it cannot be unique and so it does not exist.
780717
Divide the coefficient of the x term by 2:  -5/2\nThen square it:  (-5/2)^2 = 25/4\nAnswer: 25/4
877049
use the slope formula.\n\ny sub 2 minus y sub 1 divided by x sub 2 minus y sub 1\n\nthen you get 1-5 divided by 0-2\n\nthen you get -4 divided by -2\n\nthe answer is 2
147502
log x goes to negative infinity\nbut with the squared it becomes positive.\nSo the limit is +infinity
974331
= ln x / x^5\nln x -> inf\nx^5 -> inf\ndifferentiate both sides (Lagrange)\n\n=> 1/x  / 5x^4\n= 1/ (5x^5)\n=> 0 \n\nOr ln x increases much slower as x gets bigger, whereas x^5 increases faster, (exponentially) as it gets bigger. polynomial wins over logarithm. \n\nThe limit is 0.
854009
lim as x approaches 0 from the positive of (ln x) is negative infinity.\nlim as x approaches 0 from the positive of (sqrt x) is 0.\n\nif the numerator goes to negative infinity and the denominator goes to 0, the limit is negative infinity. it might look like L'hospital's rule is required, but it is not.
744242
let y=4x(x+x^1/2)^1/2\nlny=ln4+lnx+1/2ln(x+x^1/2)\ny'/y=1/x+{1/(x+x^1/2)}*(1+1/2x^1/2)\ny'=y[1/x+{1/(x+x^1/2)}*(1+1/2x^1/2)]
518111
As x->0, -3x->0. \nNow let -3x = z.\nThus the limit becomes\nLim z->0 [-3*{ln(1+z)}/z]\n= (-3)*Lim z->0 [{ln(1+z)}/z]\n= (-3)*1\n= -3\n\nSo 'c' is correct.
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 ? :))
986332
y=ln[sec5x+tan5x]\ny'={1/[sec5x+tan5x]}[5sec5xtan5x+5sec^25x]\n=5sec5x[sec5x+tan5x]/[sec5x+tan5x]\n=5sec5x
742119
asymptote at x=5 and y=2
1031270
x^2 + 5 + 5x + x y +5y\n\nx^2 + xy + 5x + 5y + 5\n\nx(x+y) + 5(x+y) + 5\n\nx(x+y) + 5(x+y+1)       OR     (x+5)(x+y)+5\n\nCynthia
714041
10x+4y+4\n\nor can be written as y= -2.5x-1
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
399118
y-int= (0,-5)\nx-int= (-1.2,0)
311134
f(x) = 1/(x+5)(x+2)\n\nso -5 and -2 are restricted from the domain since then we would have the denominator = 0 and the function undefined.
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.