<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>
B. angle 2 and angle 3 and congruent.
Please press brainliest if this helped you.
Answer:
x = 32
Step-by-step explanation:
∠BCA = ∠DBA (90 - ∠DBC)
∠A = ∠A
ΔABD similar to ΔACB
AC/AB = AB/AD
x / 8 = 8 / 2
x = 32
