Slip eternal,
You and I must have been working on this problem at the same time. I did it differently
lim [e^{x}+x]^{9/x}
x->infinity
=lim e^{x}(1+(x/e^{x} )]^{9/x}
x->infinity
=lim e^{x}^{9/x}(1+(x/e^{x} )]^{9/x}
x->infinity
=lim e^{9}(1+(x/e^{x} )]^{9/x}
x->infinity
= e^{9}*1
= e^{9 Extra Little Note: SlipEternal turned it into a algebraic fraction so she/he could use L'hopital's Rule to solve it. }L'Hôpital's rule - Wikipedia, the free encyclopedia
It didn't evaluate to 1 from the get go because you cannot plug in infinity. Infinity is not a number. is another way of writing . I used that notation because I wanted to be clear what was in the exponent of e. So, I did . That was how I started. Then, using the properties of natural log, the exponent becomes a coefficient. Since is a composition of continuous functions, I used the limit law that states that if is a continuous function, then . That's how I brought the limit inside the function. Then, on line 2, the expression is approaching as , so we can apply L'Hospital's Rule. I took the derivative of the top and the derivative of the bottom. Then, I skipped a step. We had . As , and . So, .
To recap, is not defined because is not a number. So, instead, I altered the form so that I could apply L'Hospital's Rule.
lim [e^{x}+x]^{9/x}
x->infinity
I have a slightly different explanation of why you can't just substitute in infinity.
e^{x}+x is infinitely large as x approaches infinity.
9/x is infinitely small as x approaches infinity.
They are not approaching infinity or 0 at the same rate. You have to change the form or it will not be valid.
Also,
Using your logic I think you might be tempted to say that
lim [e^{x}]^{9/x} = 1
x->infinity
but you can see that it can be transformed into
lim e^{9x/x}
x->infinity
lim e^{9}
x->infinity
= e^{9}
Think it over, I hope that it will help.