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- Thread starter redrage
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Please show us what you have tried andView attachment 27290

Anyone know how to solve this? Recently goes this as a question and kept getting stuck after factoring x from the bottom.

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Please share your work/thoughts about this problem

Heres what i've tried, I have a feeling I did something wrong and that it isn't actually undefined.

Because once I insert the limit then 3((e^x)-1)/x = 3 because ((e^x)-1)/x=1

However 2x+9x^2 = 2(0)+9(0)^2

So all of it will be (7+3(1))*(1/0) which will give me undefined.

Because once I insert the limit then 3((e^x)-1)/x = 3 because ((e^x)-1)/x=1

However 2x+9x^2 = 2(0)+9(0)^2

So all of it will be (7+3(1))*(1/0) which will give me undefined.

Last edited:

We haven't gotten this far into our lectures, we've just been applying limits.Look at that again. Both the numerator and denominator go to 0 when we take the limit. It may be undefined and it may not be.

Do you know how to use l'Hopital's rule?

-Dan

Also sorry i didn't exactly make it clear but we're applying this limit rule (((e^x)-1)/x) = 1

Last edited:

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This is fine! Remember, lim (a+b) = lim a + lim b, and the same works for products. Cancel out the x, use the e^x limit, and bring it all together in the end.Heres what i've tried, I have a feeling I did something wrong and that it isn't actually undefined.

View attachment 27298

Because once I insert the limit then 3((e^x)-1)/x = 3 because ((e^x)-1)/x=1

However 2x+9x^2 = 2(0)+9(0)^2

So all of it will be (7+3(1))*(1/0) which will give me undefined.

- Joined
- Jan 29, 2005

- Messages
- 11,140

Please consider the graph: LOOK HEREView attachment 27290

Anyone know how to solve this? Recently goes this as a question and kept getting stuck after factoring x from the bottom.