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

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

zepdrix · Answer

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?

whpalmer4 · Answer

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!