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anonymous
 5 years ago
find the limits when h >0 h /tan^2h
anonymous
 5 years ago
find the limits when h >0 h /tan^2h

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0h approaches 0 and tan^2h approaches 0 as h approaches 0, so you have an indeterminate form. Use L'Hopital's rule to obtain\[\lim_{h \rightarrow 0^+}\frac{h}{\tan^2h}=\frac{\lim_{h \rightarrow 0^+}h}{\lim_{h \rightarrow 0^+}\tan^2h}=\frac{\lim_{h \rightarrow 0^+}1}{\lim_{h \rightarrow 0^+}2\tan (h) \sec^2 h}=\frac{1}{\lim_{h \rightarrow 0^+}2\tan (h) \sec^2 h}\rightarrow +\infty\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Refresh the screen if it doesn't show up properly.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That's the limit from the right (i.e. heading toward 0 from the positive part of the number line).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0From the derivative of tan(x):\[\frac{d}{dx}\tan^2x=2\tan x \sec ^2 x\](chain rule)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i didnt know u take the derivative

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{h \rightarrow 0^}\frac{h}{\tan^2h}=\frac{\lim_{h \rightarrow 0^}h}{\lim_{h \rightarrow 0^}\tan^2h}=\frac{\lim_{h \rightarrow 0^}1}{\lim_{h \rightarrow 0^}2\tan (h) \sec^2 h}=\frac{1}{\lim_{h \rightarrow 0^}2\tan (h) \sec^2 h}\rightarrow \infty\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You take the derivative for L'Hopital's rule.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0In its original form, you can't say anything about the limit. Anything where you have\[\frac{0}{0}, \pm \frac{\infty}{\infty}\]is an indeterminate form (you can't say anything about it). We use (usually) L'Hopital's rule to reduce it into a determinable form.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0when u cant do anything about the limit

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If you look above, I tell you when you have an indeterminate form.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Here, the h in the numerator approached 0 the tan^2h in the denominator approached 0 when h approached 0, so you had a case of 0/0 which is indeterminate.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You look at the numerator and denominator separately first (i.e. to see if you get an indeterminable form). If it's the case, apply L'Hopital's rule or something else (but no one uses the 'something else').

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanks.. very good explanation
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