<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 second table of values.
Step-by-step explanation:
Let's put the x-values in the second table of values in correct number order:
x: -3, -2, -1, 0, 1
Now, let's write out the y-values in correct number order:
y: 1/4, 1, 4, 16, 64
Finally, let's rewrite the second table of values with the x-values in order and the corresponding y-values underneathe:
x: -3, -2, -1 0 1
y: 64, 16, 4, 1, 1/4
As it can be seen, as the x-values get bigger in value, the y-values get smaller exponentially, which is the definition of exponential decay.
Answer:
yes
Step-by-step explanation:
Answer:
0.7
Step-by-step explanation:
