Question

How do you find the limit of `(x^2+5x-7)/(e^(2x))` as x approaches `oo`?

Answer 1

`=0`

Explanation:

`lim_{x to oo} (x^2+5x-7)/(e^(2x))`

by inspection, the dominant term is in the denominator, ie the exponent, so the answer seems obvious, but we can do some more stuff to explore the obvious.....!

the immediate limit is `oo/oo` ie indeterminate so we can use L'Hopitals Rule

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`\lim _{x\to c} \frac {f(x)}{g(x)}=`

`\lim _{x\to c}\frac {f'(x)}{g'(x) }`

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`lim_{x to oo} (x^2+5x-7)/(e^(2x))`

`= lim_{x to oo} (2x+5)/(2e^(2x))`

`= lim_{x to oo} (2)/(4e^(2x))`

`= (2)/ {lim_{x to oo} (4e^(2x))}`

`= 2/oo`

`=0`