anonymous
  • anonymous
find the limits when h ->0 h /tan^2h
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
h 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
  • anonymous
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anonymous
  • anonymous
where did u get sec

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anonymous
  • anonymous
That's the limit from the right (i.e. heading toward 0 from the positive part of the number line).
anonymous
  • anonymous
From the derivative of tan(x):\[\frac{d}{dx}\tan^2x=2\tan x \sec ^2 x\](chain rule)
anonymous
  • anonymous
i didnt know u take the derivative
anonymous
  • anonymous
\[\lim_{h \rightarrow 0^-}\frac{h}{\tan^2h}=\frac{\lim_{h \rightarrow 0^-}h}{\lim_{h \rightarrow 0^-}\tan^2h}=\frac{\lim_{h \rightarrow 0^-}1}{\lim_{h \rightarrow 0^-}2\tan (h) \sec^2 h}=\frac{1}{\lim_{h \rightarrow 0^-}2\tan (h) \sec^2 h}\rightarrow -\infty\]
anonymous
  • anonymous
You take the derivative for L'Hopital's rule.
anonymous
  • anonymous
In 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
  • anonymous
Here: http://en.wikipedia.org/wiki/L'H%C3%B4pital's_rule
anonymous
  • anonymous
ohhhh,, nice
anonymous
  • anonymous
so when u do that?
anonymous
  • anonymous
when u cant do anything about the limit
anonymous
  • anonymous
If you look above, I tell you when you have an indeterminate form.
anonymous
  • anonymous
0/0 +/ infty/infty
anonymous
  • anonymous
ok
anonymous
  • anonymous
Here, 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
  • anonymous
You 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
  • anonymous
jajjaaj
anonymous
  • anonymous
ok
anonymous
  • anonymous
:)
anonymous
  • anonymous
thanks.. very good explanation
anonymous
  • anonymous
welcome

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