The Student Room Group

Limits question help

IMG_1794.jpgCan someone help me with 3(iv)?
Divide through by x as a first step...
Reply 2
Original post by DFranklin
Divide through by x as a first step...


wouldn't you move ax to the left so that e^x - ax=0, and then differentiate?
thats what I did but I don't know what to do next
Original post by SWISH99
wouldn't you move ax to the left so that e^x - ax=0, and then differentiate?No I wouldn't. Why do you think that would help?
Reply 4
Original post by DFranklin
No I wouldn't. Why do you think that would help?


thats what I did for question ii)
Original post by SWISH99
thats what I did for question ii)
Yes, and when you did it for (ii), you got e^x - 1, which you deduced was >=0 for x >=0. However, note that e^x - 1 equals 0 when x = 0. So this is already the best you can do using this simple method - if you take α>1\alpha > 1 then ddx(exαx)=exα\dfrac{d}{dx}(e^x - \alpha x) = e^x - \alpha which is < 0 when x = 0.
Reply 6
Original post by DFranklin
Yes, and when you did it for (ii), you got e^x - 1, which you deduced was >=0 for x >=0. However, note that e^x - 1 equals 0 when x = 0. So this is already the best you can do using this simple method - if you take α>1\alpha > 1 then ddx(exαx)=exα\dfrac{d}{dx}(e^x - \alpha x) = e^x - \alpha which is < 0 when x = 0.


I see what you're saying. So how would dividing by x help?
what year are yo in?
Original post by SWISH99
I see what you're saying. So how would dividing by x help?
Have a think about what "the biggest value α\alpha s.t. exxα\dfrac{e^x}{x} \geq \alpha for x >=0" means in terms of the graph of exx\dfrac{e^x}{x}.

(I don't want to spell this out for you, this is the kind of thing you need to get used to understanding on your own).
Reply 9
Original post by SalvadorMendes
what year are yo in?
why?
Reply 10
Original post by DFranklin
Have a think about what "the biggest value α\alpha s.t. exxα\dfrac{e^x}{x} \geq \alpha for x >=0" means in terms of the graph of exx\dfrac{e^x}{x}.

(I don't want to spell this out for you, this is the kind of thing you need to get used to understanding on your own).


Oh I think I get what you mean. Differentiating e^x/x in order to find the minimum point?
Original post by SWISH99
Oh I think I get what you mean. Differentiating e^x/x in order to find the minimum point?
Yes, that's right :smile:
this is complicated and I'm in year 10 and I'm getting really scared
Original post by SalvadorMendes
this is complicated and I'm in year 10 and I'm getting really scared


This is undergrad, you won't have to worry about limits for a while. (the earliest you'll encounter them is probably year 13)
oh thank god:u:

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