limit at infinity of { [e^(2x+1)] / (e^2x) + 1}

[@saask_me]

lim(x->-

){[e^(2x+1)]/(e^2x)+1}

 

[Deleted member 4993]

lim(x->-

){[e^(2x+1)]/(e^2x)+1}

As posted the function looks like:

\(\displaystyle \dfrac{e^{2x+1}}{e^{2x}} \ + \ 1\)

Is that correct?

 

[@saask_me]

As posted the function looks like:

\(\displaystyle \dfrac{e^{2x+1}}{e^{2x}+1}\)

Is that correct?

I'm sorry but that " + 1 " is included in the denominator. Can you help me?

 

[stapel]

I'm sorry but that " + 1 " is included in the denominator.

\(\displaystyle \dfrac{e^{2x+1}}{e^{2x}+1}\)

Can you help me?

You can use basic algebra to try to break things apart a little:

. . . . .\(\displaystyle e^{2x+1}\, =\, \left(e^{2x}\right)\, \left(e^1\right)\)

What happens if you use the standard step of dividing through, top and bottom, by the leading term of e^2x?

 

[pka]

lim(x->-

){[e^(2x+1)]/(e^2x)+1)}

Divide both numerator and denominator by \(\displaystyle e^{2x}\).