<span>If you plug in 0, you get the indeterminate form 0/0. You can, therefore, apply L'Hopital's Rule to get the limit as h approaches 0 of e^(2+h),
which is just e^2.
</span><span><span><span>[e^(<span>2+h) </span></span>− <span>e^2]/</span></span>h </span>= [<span><span><span>e^2</span>(<span>e^h</span>−1)]/</span>h
</span><span>so in the limit, as h goes to 0, you'll notice that the numerator and denominator each go to zero (e^h goes to 1, and so e^h-1 goes to zero). This means the form is 'indeterminate' (here, 0/0), so we may use L'Hoptial's rule:
</span><span>
=<span>e^2</span></span>
Answer:
the answer is 4.94 because you subtract 10.93-5.99
I think it’s 84 you multiply 14 and 12 then divide by 2 since it is a triangular prism
Answer:56
Step-by-step explanation:
with 16 into 55 do that math then put the number you got then put it into another 16 boom 56
Answer:
<em>Choice: B.</em>
Step-by-step explanation:
<u>Operations With Functions</u>
Given the functions:
![f(x)=\sqrt[3]{12x+1}+4](https://tex.z-dn.net/?f=f%28x%29%3D%5Csqrt%5B3%5D%7B12x%2B1%7D%2B4)

The function (g-f)(x) can be obtained by replacing both functions and subtracting them as follows:

![(g-f)(x)= \log(x-3)+6 - (\sqrt[3]{12x+1}+4)](https://tex.z-dn.net/?f=%28g-f%29%28x%29%3D%20%5Clog%28x-3%29%2B6%20-%20%28%5Csqrt%5B3%5D%7B12x%2B1%7D%2B4%29)
Operating:
![(g-f)(x)= \log(x-3)+6 - \sqrt[3]{12x+1}-4](https://tex.z-dn.net/?f=%28g-f%29%28x%29%3D%20%5Clog%28x-3%29%2B6%20-%20%5Csqrt%5B3%5D%7B12x%2B1%7D-4)
Joining like terms:
![\boxed{(g-f)(x)= \log(x-3) - \sqrt[3]{12x+1}+2}](https://tex.z-dn.net/?f=%5Cboxed%7B%28g-f%29%28x%29%3D%20%5Clog%28x-3%29%20-%20%5Csqrt%5B3%5D%7B12x%2B1%7D%2B2%7D)
Choice: B.