If you're going to use L'Hospital's Rule
by L'Hospital's Rule
by L'Hospital's Rule
.
I need help with this question. I think I'm supposed to use L'Hopital's Rule, but I'm not sure if I'm doing it correctly.
lim as x -->0 (2x^6+6x^3)/(4x^5+3x^3)
I got 0 at first, but then I got 2. I'm just so unsure! Any help at all is greatly appreciated!
I applied L'Hopital's rule like 4 times then stopped. I knew that would be wrong :/
But how did you figure out this part?
lim as x-->0= 2x^3+ 6/ 4x^2+3
I thought, you just went straight to finding the derivative after the original was found to be indeterminate? Please tell me how that was found.
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We do get rid of the x's !!
Now it's safe to set the remaining ones all to zero as we are beyond getting 0/0
which was the indeterminate form.
As you were asked to apply L'Hopital's Rule...
Initially, we have 0/0, so...
Differentiate numerator and denominator
Still 0/0, so differentiate again...
Still 0/0, so differentiate again..
It's no longer indeterminate, especially there is no potential for zero in the denominator,
so now it's safe to set the remaining x's to zero to evaluate the limit.