Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

differential equations

How to do these types of questions where a value becomes large?

Part 2 of question 3
Please anyone?

(edited 4 years ago)

OP

Whats the logic behind this question

(edited 4 years ago)

Original post by Vand3rwaal

How to do these types of questions where a value becomes large?

Part 2 of question 3
Please anyone?

Upload your solution to part (i)

OP

Original post by vbzl

Upload your solution to part (i)

image-40503dc3-1766-4c5a-a799-aed617767cc02376373087972787982-compressed.jpg.jpeg

Original post by Vand3rwaal

image-40503dc3-1766-4c5a-a799-aed617767cc02376373087972787982-compressed.jpg.jpeg

As t becomes very large, what happens to 2/(1+t)?

OP

Idk

Original post by Vand3rwaal

Idk

Ok let's take an example.
y=1/x
When x =10, y=1/10=0.1
When x is large say 10000, y=1/10000=0.0001
So can you figure out now what happens to 2/(1+t) as t becomes very large?
You may try any large value on your calculator and see what happens.

OP

It becomes closer and closer to zero

Original post by Vand3rwaal

It becomes closer and closer to zero

Right, and the same for

\dfrac{2}{1 + t}

. Who cares what's on the top, eventually as t gets large enough it goes to 0 because you're dividing a smaller number by a much large one.

So when you've got

e^{\frac{2}{t+1} - 2}

, what value does the exponent approach?

(edited 4 years ago)

OP

100e^-2

OP

Yeah thats the answer! 😅
I get it now
Thanks alot!

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