Verified answer

you have to think of e^(2x)-1 and x as two diff functions if they both approach infinity on the y axis as the x values go to infinity then you can apply lhopitals rule

since e is just another constant when the x of e^x go to infinity so does the whole thing

x is obviously going to infinity so you can apply lhopitals

after you apply lhopitals you get [2e^(2x)]/1

which goes to infinity (you can look at the graph for proof)

if lim x->0 for f(x), this limit will become an indeterminate form

This means we have to use L'Hopital's Rule.

Which tells us to take derivative of top and bottom and evaluated it again.

lim x-> 0 of 2e^(2x)= 2.

lim(x->0) [e^(2x) - 1] /x => 0/0 , indet. apply. L'Hospital's rule:

= lim(x->0) d[e^(2x) - 1]/dx / (dx)/dx

= lim(x->0) 2e^(2x)

= 2 * e^(0)

= 2 * 1

= 2