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Please help with these questions-limits etc.

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Original post by betapro
Yes for 2 i got 6) so L=1 and not a clue how to do 4 :biggrin:


Just by looking at Q4 I'd say I would test a base case of n=1n=1 before evaluating what happens as nn \rightarrow \infty as see if the inequality still holds, examining closely the denominators if two things tend to the same thing. Just a thought though.
Reply 21
For problem 4 i got my ansdwer as (a) is correct
Original post by betapro
Found values for which absolute values vanish and split into three intervals and then tested each one


I got a different result when x is less than -1/2.
Original post by betapro
For problem 4 i got my ansdwer as (a) is correct


What was wrong with (e)?
Reply 24
Original post by EricPiphany
What was wrong with (e)?


For e i agree with first step but i dont get how u get that to 1
Reply 25
Original post by EricPiphany
I got a different result when x is less than -1/2.


Could you send screenshot of working to see where i went wrong? ty <3
Original post by betapro
For e i agree with first step but i dont get how u get that to 1


You can always use a little cheeky algebra :wink:
Original post by betapro
Could you send screenshot of working to see where i went wrong? ty <3


For x<1/2x<-1/2, both 2x+1,2x1<02x+1, 2x-1 < 0, so LHS is (2x+1)(2x1)=22(-2x+1)-(-2x-1)=2 \le 2...
Unless I've done something really dumb somewhere...
Reply 28
Original post by EricPiphany
For x<1/2x<-1/2, both 2x+1,2x1<02x+1, 2x-1 < 0, so LHS is (2x+1)(2x1)=22(-2x+1)-(-2x-1)=2 \le 2...
Unless I've done something really dumb somewhere...


Right yea ur right so if you would say that the answer is 6?
Reply 29
Original post by EricPiphany
You can always use a little cheeky algebra :wink:


Can u show me this hidden magic of algebra xD
Original post by betapro
Right yea ur right so if you would say that the answer is 6?


I'll let you decide on that one...
Reply 31
Original post by EricPiphany
I'll let you decide on that one...


Well for most right interval u get -2 less than equal to 2 so u take that to be right, and from previous working we know that -infinity to -0.5 is right and also the interval between -0.5 and half is also right then yes? A confirmation would be great :biggrin:
Original post by betapro
Can u show me this hidden magic of algebra xD


consider 2n3n+3=23×nn+1 \dfrac{2n}{3n+3} = \dfrac{2}{3} \times \dfrac{n}{n+1}...
or simply multiply by the denominator...
Original post by betapro
Well for most right interval u get -2 less than equal to 2 so u take that to be right, and from previous working we know that -infinity to -0.5 is right and also the interval between -0.5 and half is also right then yes? A confirmation would be great :biggrin:


https://www.desmos.com/calculator/7scpx7ack1
Reply 34


Graph doesnt break or anything so i see why its -infitinty to positive infinity thanks. Still requiring some further assistance with that question 4 with the getting to one situation xD
Original post by betapro
Graph doesnt break or anything so i see why its -infitinty to positive infinity thanks. Still requiring some further assistance with that question 4 with the getting to one situation xD


Original post by EricPiphany
consider 2n3n+3=23×nn+1 \dfrac{2n}{3n+3} = \dfrac{2}{3} \times \dfrac{n}{n+1}...
or simply multiply by the denominator...


nn+1\dfrac{n}{n+1} is clearly less than 1, so 23×nn+1<23\dfrac{2}{3} \times \dfrac{n}{n+1} < \dfrac{2}{3}...
Reply 36
Original post by EricPiphany
nn+1\dfrac{n}{n+1} is clearly less than 1, so 23×nn+1<23\dfrac{2}{3} \times \dfrac{n}{n+1} < \dfrac{2}{3}...


Yea see what u mean but surely it could never be 1, or is that just the notation you are meant to use?
Original post by betapro
Yea see what u mean but surely it could never be 1, or is that just the notation you are meant to use?


\le means less than or equal to.
It is true if << or == or both.

so if x<yx<y then xyx \le y
Reply 38
Original post by EricPiphany
\le means less than or equal to.
It is true if << or == or both.

so if x<yx<y then xyx \le y


Derp yea...

Thanks so much for your help, just to conclude for answers:
q1-3
q2-6
q3-4
q4-2
q5-2
q6-6

Also is there a way i can give u rating? <3
Original post by betapro
Derp yea...

Thanks so much for your help, just to conclude for answers:
q1-3
q2-6
q3-4
q4-2
q5-2
q6-6

Also is there a way i can give u rating? <3


i can't confirm them

and not really, but you can click the green thumbs up under a post by me

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