Question

The limit as h approaches 0 of (e^(2+h)-e^2)/h = ?
The answer is e^2; please explain how to get that?

Answer 1

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.

Answer 2

Notice,\[\frac{e^{2+h}-e^2}{h}=\frac{e^2(e^{h}-1)}{h}\]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:\[\lim_{h \rightarrow 0}\frac{e^2(e^{h}-1)}{h}=\frac{\lim_{h \rightarrow 0}e^2\lim_{h \rightarrow 0}(e^{h}-1)}{\lim_{h \rightarrow 0}h}=\frac{e^2\lim_{h \rightarrow 0}(e^h-1)'}{\lim_{h \rightarrow 0}h'}=\frac{e^2\lim_{h \rightarrow 0}e^h}{\lim_{h \rightarrow 0}1}\]\[=\frac{e^2.1}{1}=e^2\]

Answer 3

When I tried L'Hopital's rule, the first thing I got was e^(2+h)-2e^2. Is that wrong?

Answer 4

How did you get that?

Answer 5

The derivative of e^2 with respect to h isn't 2e^2, it's 0.

Answer 6

Whoops. Thanks I see it now. I mixed my rules up.