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C4 Math question

AS t→infinity, x tends to a limiting value, find this limiting value
the equation is (2e^t -2)/(2e^t -1)
can someone explain to me what this means
thanks

(edited 7 years ago)

Reply 1

RDKGames

Study Forum Helper

Original post by mlyke

AS t→infinity, x tends to a limiting value, find this limiting value
the equation is (2e^0.5 -2)/(2e^0.5 -1)
can someone explain to me what this means
thanks

That's not an equation, that's a number....

Reply 2

mlyke

Original post by RDKGames

That's not an equation, that's a number....

oh oops,
replace the 0.5 with ts

Reply 3

Mr M

Study Forum Helper

Original post by mlyke

AS t→infinity, x tends to a limiting value, find this limiting value
the equation is (2e^t -2)/(2e^t -1)
can someone explain to me what this means
thanks

Note that

\displaystyle \frac{2e^t - 2}{2e^t -1} = \frac {2e^t -1 - 1}{2e^t - 1} = 1 - \frac{1}{2e^t-1}

Reply 4

mlyke

Original post by Mr M

Note that

\displaystyle \frac{2e^t - 2}{2e^t -1} = \frac {2e^t -1 - 1}{2e^t - 1} = 1 - \frac{1}{2e^t-1}

could you explain that, haven't learn that in class and would like to understand it
thanks

Reply 5

Mr M

Study Forum Helper

Original post by mlyke

could you explain that, haven't learn that in class and would like to understand it
thanks

The middle bit just says

-2 = -1 -1

so I'm sure you follow that.

The bit on the right just uses the fact that a number divided by itself = 1 (this is true for all numbers except 0).

Reply 6

mlyke

Original post by Mr M

The middle bit just says

-2 = a-1 -1

so I'm sure you follow that.

The bit on the right just uses the fact that a number divided by itself = 1 (this is true for all numbers except 0).

Mh i understand that but how does infinity play a role here, sorry if this sounds a bit ignorant just want to understand the thought process
The answer is 1 so its obviously right

Reply 7

RDKGames

Study Forum Helper

Original post by mlyke

oh oops,
replace the 0.5 with ts

Okay so you have

\displaystyle x=\frac{2e^t-2}{2e^t-1}

. Now you can do what Mr M has suggested. Otherwise, an alternative is to divide top and bottom of the fraction by

e^t

so it becomes

x=\frac{2-2e^{-t}}{2-e^{-t}}

and as

t\rightarrow \infty

you have

e^{-t} \rightarrow 0

then what does x tends to?

Reply 8

Mr M

Study Forum Helper

Original post by mlyke

Mh i understand that but how does infinity play a role here, sorry if this sounds a bit ignorant just want to understand the thought process
The answer is 1 so its obviously right