# How to solve this limit?

#### redrage

##### New member

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

#### Subhotosh Khan

##### Super Moderator
Staff member
View attachment 27290
Anyone know how to solve this? Recently goes this as a question and kept getting stuck after factoring x from the bottom.
Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

#### redrage

##### New member
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.

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#### redrage

##### New member
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.

#### topsquark

##### Senior Member
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

#### redrage

##### New member
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
We haven't gotten this far into our lectures, we've just been applying limits.

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

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#### Zermelo

##### Junior Member
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.
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.