Limit

[TheNextOne]

Find the following limit:

lim ( as x approaches infinity) [6e^2x - e^x + 7e^-x]/ [3e^2x + 1]

How should I solve this? [/img]

 

[skeeter]

\(\displaystyle \frac{6e^{2x} - e^x + 7e^{-x}}{3e^{2x} + 1}\)

divide each term by \(\displaystyle e^{2x}\) ...

\(\displaystyle \frac{6 - e^{-x} + 7e^{-3x}}{3 + e^{-2x}}\)

now take the limit as x->infinity.

 

[TheNextOne]

Is the answer 3?

 

[stapel]

Is the answer 3?

How did you get this answer?

Please reply showing all of your work and reasoning. Thank you.

Eliz.

 

[TheNextOne]

Answer = 3

Well, if i divide everything by e^2x and use infinity as the limit, then all of the e^-x terms become e^0 = 1.

The top will be 6-1+7 and the bottom 3+1.

Thus 12/4= 3

 

[skeeter]

no ...

all the e^(-kx) terms go to 0 as x->infinity, not 1.

the limit is 2.

 

[tkhunny]

Re: Answer = 3

all of the e^-x terms become e^0

Try that again. How does x increasing without bound suddenly become zero?

 

[crayzeerunner]

Cant you also look at the coefficients of the highest powered leading term and just divide them. 6/3 = 2?