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

This Question is Closed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Notice,\[\frac{e^{2+h}e^2}{h}=\frac{e^2(e^{h}1)}{h}\]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^h1 goes to zero). This means the form is 'indeterminate' (here, 0/0), so we may use L'Hoptial's rule:\[\lim_{h \rightarrow 0}\frac{e^2(e^{h}1)}{h}=\frac{\lim_{h \rightarrow 0}e^2\lim_{h \rightarrow 0}(e^{h}1)}{\lim_{h \rightarrow 0}h}=\frac{e^2\lim_{h \rightarrow 0}(e^h1)'}{\lim_{h \rightarrow 0}h'}=\frac{e^2\lim_{h \rightarrow 0}e^h}{\lim_{h \rightarrow 0}1}\]\[=\frac{e^2.1}{1}=e^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0When I tried L'Hopital's rule, the first thing I got was e^(2+h)2e^2. Is that wrong?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0How did you get that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The derivative of e^2 with respect to h isn't 2e^2, it's 0.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Whoops. Thanks I see it now. I mixed my rules up.
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