The limit as h approaches 0 of (e^(2+h)-e^2)/h is ?

1 answer:

3 0

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.
[e^(2+h) − e^2]/h = [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:

=e^2

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