Use a calculator to estimate the values of the following limits to two decimal places. lim h → 0 (2.7^h − 1)/h lim h → 0 (2.8^h − 1)/h
you don't need to use a calculator for this problem, you need to do is differentiate with the top and the bottom with respect to h
see at your level anything divided by zero is equal to infinite, so you use a rule called the "L'Hospital's rule", which indicates that you may differentiate the numerator and the denominator
you can also use in your differentiation the D operator if you kindly know what it is?
so the answer to the first part would be 1.0296... and for 2 decimal places 1.02
can you work out the other one yourself, or if you need my help i will do
ok let me prove it for you
recall \[a ^{x}, f' = a ^{x}\ln(a)\]
now in your case because a=2.8 and the limit ->0, so all you need to do is to use your calculator to take the natural log of 2.8 or 2.7
because anything to the power of zero is one
for 2.7 limit the answer would be 0.99
i hope that makes sense to you
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