Lim x (is going to 0) x/(x^2 + 2x)= Lim x (is going to 0) x/[x(x + 2)]= Lim x (is going to 0) 1/(x + 2) = 1/2
Lim x (is going to 0) x/(x^2 + 2x)= Lim x (is going to 0) x/[x(x + 2)]= Lim x (is going to 0) 1/(x + 2) = 1/2
The derivative of a constant is always 0. To show this, let's apply the definition of derivative. Recall that the definition of derivative is: f'(x) = lim h→0 (f(x + h) - f(x))/h Let f(x) = 1. Then: f'(x) = lim h→0 (1 - 1)/h = lim h→0 0/h = lim h→0 0 = 0!
lim(x->0) of sin(x)^2/x we use L'Hospital's Rule and derive the top and the bottomd/dx(sin(x)^2/x)=2*sin(x)*cos(x)/1lim(x->0) of 2*sin(x)*cos(x)=2*0*1=0
Using calculus to see if the function f(x) is continuous at a point (point c) involves three steps. These three conditions must be met: 1. f(c) exists, is defined 2. lim f(x) exists x-->c 3. f(c)= lim f(x) x-->c
Unfortunately, the browser used by Answers.com for posting questions is incapable of accepting mathematical symbols. This means that we cannot see the mathematically critical parts of the question. We are, therefore unable to determine what exactly the question is about and so cannot give a proper answer to your question. Please resubmit your question spelling out the symbols as "plus", "minus", "times", "divided by", "equals" etc. As it appears, you seem to be seeking the limit of sin(4x)*sin(6x) as x tends to 0. Both components of the product tend to 0 as x tens to 0 and so the limit is 0. Bit I suspect that is not the limit that you are looking for.
Lim x (is going to 0) x/(x^2 + 2x)= Lim x (is going to 0) x/[x(x + 2)]= Lim x (is going to 0) 1/(x + 2) = 1/2
For ease of writing, every time I say lim I mean the limit of _____ as x approaches zero. Lim sin3xsin5x/x^2=lim{[3sin(3x)/(3x)][5sin(5x)/(5x)]}. One property of limits is that lim sin(something)/(that same something)=1. So we now have lim{[3(1)][5(1)]}=15. lim=15
If you mean: Lim(x→2) 3x + 1 Then the answer is 7 If you mean something else, then you'll need to clarify your question.
yes
size_t trim (char *p) { char *from, *lim, *to; from= p; lim= p + strlen (p); to= p; while (lim>from && lim[-1]==' ') --lim; while (from<lim && from[0]==' ') ++ from; while (from<lim) *to++ = *from++; *to= '\0'; return to-p; }
lim x -> -inf [x/ex] = lim x -> +inf[-x/e-x] = - lim x -> +inf [ xex ] = -infIf you want to see this function then I suggest you use either:(a) wolframalpha.com: put in show me x/exp(x)or (b) geogebra, which is available for the desktop.
The surname Lim comes from China, it was told that a baby of an emperor was born in the "Woods" (Lim in Chinese), the baby was given the surname Lim.
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Marcus Lim's birth name is Marcus Lim Jiing-Xian.
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The reality is this, Mr Lim Yeow Hua @ Lim You Qin...Lim would be the last name, maybe the Yeow Hua is the first name, but how about the @ Lim You Qin, is it an alias name? please help....
Michael Kho Lim's birth name is Michael Kho Lim.