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Let f(x)=e^x Clearly: (e^x - 1)/x = (e^x - 1)/(x-0) = ( f(x) - f(0) )/x-0 ----> f'(0) as x--> 0

0^0 is undefined.

Thanks.

Homework Statement Let f(x)=e^(-x)-x ,, x belongs to R Find the domain of f inverse Homework Equations Domain of f inverse = range of f The Attempt at a Solution we have : -inf < x < inf -inf < -x < inf ... (1) 0 < e^(-x) < inf ... (2) By adding (1) and (2) ...

ohhhh sorry :| I finished it thanks <3

Homework Statement Solve the following equation for x: e^{3x}+sinh(x)=0 Homework Equations None. The Attempt at a Solution It the same as: e^{3x}+\frac{e^{2x}}{2}-\frac{e^{-2x}}{2}=0 Multiply by 2 2e^{3x}+e^{2x}-e^{-2x}=0 Multiply by e^(2x) 2e^{5x}+e^{4x}=1...

Its better to post your work. Here is a start: the integration of 1/(3x+y)^2 with respect to y from y=3 to y=4 is : (-1/3x+4) - (1/3x+3) and this result is easy to integrate with repsect to x, isn't it ?

You have to use the cauchy's test? If not: For an easier life, use the integral test.

Its a sequence or a series ? There is no comparison test for sequences. If its a sequence: Try the sandwich theorem. If its a series: See Dick's reply.

(2+e^(-x))/x > 2/x

Another one: put t=n+1 Clearly, t goes to infinity as n goes to infinity, so the limit will be : lim of (t-1)/t^3 as t goes to infinity Devide top and bottom by t: lim of (1 - (1/t) )/t^2 as t goes to infinity = (1-0)/infinity=1/infinity=0

If f(n)<g(n) for n>a then: f(n)<g(n) for n>a+1

If you mean less than or EQUAL , then you could consider the inequality starts from n=2, and this will be remove the "EQUAL" sign from the inequality Finally, you know that the finite sum does not affect the convergence of the series, so it will include the 1.

your website said: if L < 1 then the series absolutely convergent (hence convergent). Its agree with my conclusion :)

by the way. the first statement is true. its one-to-one function.