L hopital rule

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.

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Can you solve it by taking the lin in the both sides 

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

I don't understand what you are asking!

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

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

In this case L'Hopital's rule makes things ten times worse!

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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 

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.   :)

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 " 

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

...

10⁶    4.99999*10^-6           5*10^-6

10¹⁰  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.

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nice 

thank you