This seems to come out to as did the limit of (as x approached infinity) But the computer says the answer is NOT 1. http://www.freemathhelp.com/forum/im...n_confused.gif

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- Nov 4th 2013, 09:04 PMJason76e Limit Problem

This seems to come out to as did the limit of (as x approached infinity) But the computer says the answer is NOT 1. http://www.freemathhelp.com/forum/im...n_confused.gif - Nov 4th 2013, 09:19 PMSlipEternalRe: e Limit Problem
- Nov 4th 2013, 10:00 PMJason76Re: e Limit Problem
But how come it didn't evaluate to 1 from the get go? Why did we proceed to the derivative and then plugin?

What is ? - Nov 5th 2013, 04:06 AMMelody2Re: e Limit Problem
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 - Nov 5th 2013, 04:17 AMSlipEternalRe: e Limit Problem
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. - Nov 5th 2013, 04:57 AMMelody2Re: e Limit Problem
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. - Nov 5th 2013, 06:29 AMSlipEternalRe: e Limit Problem
- Nov 5th 2013, 03:43 PMMelody2Re: e Limit Problem
Thanks SlipEternal,

how do you write it out neatly like that? - Nov 5th 2013, 04:17 PMSlipEternalRe: e Limit Problem
- Nov 5th 2013, 04:54 PMMelody2Re: e Limit Problem
I have just down loaded Latex but I haven't tried to use it yet. I have questions but I probably need to play with it first.

Thanks