# Thread: limits using l'hospital's formula?

1. ## limits using l'hospital's formula? (SOLVED!! THANKS!!)

i'm really struggling with l'hospital's formula, and i'll get help in person but this homework is due tonight.

limit of ((e^x)-1)/sin(7x) as x --> 0

limit of ((11^x)-(8^x))/x as x --> 0

limit of ((e^x)+2x-1)/(7x) as x --> 0

sorry i'm new here, this is my first time posting so hopefully this works! thanks to anyone who can help!!

2. Basically, you keep taking the derivative of the numerator and the denominator until you can plug in the value for the limit.

For the first one, for example:

limit of ((e^x)-1)/sin(7x) as x --> 0

take derivatives of (e^x)-1 (which is e^x) and sin(7x) (which is (7cos(7x))

so now it looks like this:

limit of (e^x)/7cos(7x) as x --> 0

you can now plug in 0 without division by 0

and you get
1/7

I hope this helps with the rest

3. $\displaystyle \lim_{x\to 0}\frac{e^{x}-1}{sin(7x)}$
Take the derivative of the top and the bottom, separately.

The derivative of $\displaystyle e^{x}-1$ is $\displaystyle e^{x}$

The derivative of $\displaystyle sin(7x)$ is $\displaystyle 7cos(7x)$

So, we get $\displaystyle \lim_{x\to 0}\frac{e^{x}}{7cos(7x)}$

Now, what do you get when you plug in x=0?.

4. thanks you guys! i figured out the first one and the last one, but i can't figure out the middle one.
cause i got (x11^(x-1)-x8^(x-1))/1 which means the numerator = 0...i dunno i don't think i did it right?

thanks for the help galactus and woohoo!
edit : ohh sorry i just read the rest of your response woohoo. so i can keep taking the derivative? what about if i continue taking the derivative for the 2nd one, the denominator becomes 0? or do i just get rid of the denominator because it is 1?

5. Originally Posted by mathishard33
i'm really struggling with l'hospital's formula, and i'll get help in person but this homework is due tonight.

limit of ((11^x)-(8^x))/x as x --> 0

sorry i'm new here, this is my first time posting so hopefully this works! thanks to anyone who can help!!
Be careful!!

$\displaystyle \frac{\,d}{\,dx}a^x=a^x\ln(a)$

So, $\displaystyle \lim_{x\to0}\frac{11^x-8^x}{x}=\lim_{x\to0}\left[11^x\ln(11)-8^x\ln(8)\right]=\dots$

6. okay so the next derivative would be
(11^(x)ln(11)^(2)+x)-(8^(x)ln(8)^(2)+x)

is that correct?
in which case the answer would be 1.425824614?

7. The derivative of 11^x is 11^x ln (11)

proof:
y=11^x
ln (y) = x ln (11)
(1/y)(dy/dx) = ln (11)
dy/dx = y ln (11)
dy/dx = 11^x ln (11)

similarly, the derivative of 8^x is 8^x ln(8)

once you have taken the derivative and are able to plug in the value...do not continue to take the derivative

if you do not plug it in the first time you can, you will get an incorrect answer

8. I think you can solve this one with the first derivative

and nothing more

9. Originally Posted by woohoo
The derivative of 11^x is 11^x ln (11)

proof:
y=11^x
ln (y) = x ln (11)
(1/y)(dy/dx) = ln (11)
dy/dx = y ln (11)
dy/dx = 11^x ln (11)

similarly, the derivative of 8^x is 8^x ln(8)

once you have taken the derivative and are able to plug in the value...do not continue to take the derivative

if you do not plug it in the first time you can, you will get an incorrect answer
wow i'm stupid i was going to take its derivative again!
since x --> 0, 11^x = 1 and 8^x = 1 so its just ln(11)-ln(8)...so .3184537311?

10. thanks a ton for your help guys!!

11. Originally Posted by mathishard33
wow i'm stupid i was going to take its derivative again!
since x --> 0, 11^x = 1 and 8^x = 1 so its just ln(11)-ln(8)...so .3184537311?
I believe it would be 11 ln (11) - 8 ln (8) = 9.741....

(check earlier posts)

12. Originally Posted by woohoo
I believe it would be 11 ln (11) - 8 ln (8) = 9.741....

(check earlier posts)
well its (11^x)(ln(11)) - (8^x)(ln(8)) and since x is approaching 0, 11^x = 1 and 8^x = 1 so its just ln(11) - ln (8)

13. Originally Posted by mathishard33
well its (11^x)(ln(11)) - (8^x)(ln(8)) and since x is approaching 0, 11^x = 1 and 8^x = 1 so its just ln(11) - ln (8)

well never mind then!

see! you do this stuff better than I do!

14. Originally Posted by woohoo
well never mind then!

see! you do this stuff better than I do!
haha no i just have been working this problem for like an hour so now its basically memorized haha