Recent content by istevenson

supposing that the f value is at the interval [M,N] for \lambda and 1-\lambda are positive,f(a) not eqal to f(b) thus \lambda f(a)+(1-\lambda)f(b) >\lambda M +(1-\lambda)M = M \lambda f(a)+(1-\lambda)f(b) <\lambda N +(1-\lambda)N =N so the value of \lambda...

make a little change and it will be much easier to prove it. step1 \frac{p}{p+q}f(a)+\frac{q}{p+q}=f(c) step2 let \frac{p}{p+q}=\lambda because p and q are all positive, so \lambda is some value at the interval (0,1) then we get \lambda f(a)+(1-\lambda)f(b)=f(c) now you can use the...

precondition,do you know what is the difference between the following two expressions: \lim_{y\rightarrow0} \frac{0}{y} \lim_{y\rightarrow0}\frac{f(y)}{y} supposing when y\rightarrow0 f(y) \rightarrow0 If you do,i will tell you that your problem is really really obviously clear.

Yes, you have seen it. actually the limit doesn`t exist,because the parameter \theta in your equation isn`t a constant.

\frac{2^n}{n!}=\frac{2}{n}\frac{2}{n-1}...\frac{2}{2} \frac{2}{1} so when n\rightarrow\infty we get \frac{2^n}{n!}<\frac{4}{n} to any \varepsilon > 0 \exists N=[\frac{4}{\varepsilon}]+1 when n>N \frac{2^n}{n!}<\frac{4}{n} <\frac{4}{[\frac{4}{\varepsilon}]+1}<\varepsilon now we get the...

man,Limit of 0/y as y tends to 0 is 0 it is very clear.

suppose x = ky^2, y \rightarrow 0 then we know this: x \rightarrow 0 also. so: \lim \frac{xy^2}{x^2+y^4}= \lim \frac{ky^4}{k^2y^4+y^4}=\lim \frac{k}{k^2+1} the limit doesn`t exit, so it`s not continuous at (0,0).

OK,let`s see it \int \frac{6 du}{3 \left( \frac{4}{3} u ^ 2 + 1 \right)} = \int \frac{2 du}{\left( \left( \frac{2}{\sqrt{3}} u \right) ^ 2 + 1 \right)} and now we use the substitution: p = \frac{2}{\sqrt{3}} u then we get this equation: u =\frac{\sqrt{3}}{2} p so: \int \frac{2 du}{\left(...

the correct answer should be this: 0 < k < 2 or k = 2 or k =0.

something was wrong in step 4. becuse \int { \frac{1} {px^2+1} dx = \frac{1} {\sqrt{p}} arctan(\sqrt{p} x) +c here p > 0. thus,the correct answer is \sqrt{3} arctan(\frac{2u} {\sqrt{3}}) +c