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#3**+10 **

xvxvxv wrote: *Can you solve it by taking the lin in the both sides*

I don't understand what you are asking!

.

Alan Nov 26, 2014

#4**+10 **

Thanks Alan - I didn't know that 'by definition' bit.

Actually you haven't used L'hopitals rule though ?

Melody Nov 27, 2014

#6**0 **

But we didn't study this method

We always try to simplify it ti give us zero by zero or infinity by infinity then using l hopital rule

xvxvxv Nov 27, 2014

#7**+15 **

Best Answer

But I'm willing to be set straight!

NB: The last expression still reduces to e^{-5}, so L'Hopital's rule still works of course, but I see it as an unnecessarily complicated approach here.

.

Alan Nov 27, 2014

#8**+10 **

Hi Alan,

Yes I did not think that L'Hopital's rule helped either.

I was just pointing out that the question requested it and it was not used. That is all. :)

Melody Nov 27, 2014

#9**0 **

Nice thank you Alan.

But can you please explain to me your method

First what do you mean by

" As t tend to infinity t+2 tends to t "

xvxvxv Nov 27, 2014

#10**+10 **

Compare 5/(t+2) with 5/t for various values of t:

**t 5/(t+2) 5/t**

1 1.67 5

10 0.42 0.5

100 0.049 0.05

1000 0.00499 0.005

...

10^{6} 4.99999*10^{-6} 5*10^{-6}

10^{10 }4.999999999*10^{-10} 5*10^{-10}

You can see that as t gets bigger, 5/(t+2) becomes closer and closer to 5/t, so it's in this sense that t+2 becomes more and more like t as t tends to infinity.

.

Alan Nov 27, 2014