Question

Estimate the limit by substituting smaller and smaller values of.Give your answer to one decimal place.
lim(h-->0) e^(1+h)-e/h

Answer 1

Let's estimate by letting \(h=.00001\). That's pretty close to zero right? :)\[\large \lim_{h \rightarrow 0}\frac{e^{1+h}-e}{h} \qquad \approx \qquad \frac{e^{1+.00001}-e}{.00001}\]
When you put that into your calculator, what do you get?

Answer 2

To get the 1 decimal place, a value of \(h=0.093\) is small enough coming in from the \(0^+\) side, but you have to go to \(h=-0.05\) to get there on the \(0_-\) side. This approach doesn't seem to converge very quickly once you've got the first few digits. Not that it matters at all for this problem!