Another trig limit q: (t^2)/(1 - (cost)^2), t going to 0
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pka
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\(\displaystyle \L
\begin{array}{l}
\lim _{x \to 0} \left( {\frac{x}{{\sin (x)}}} \right) = \left[ {\lim _{x \to 0} \left( {\frac{{\sin (x)}}{x}} \right)} \right]^{ - 1} = 1 \\
\frac{{t^2 }}{{1 - \cos ^2 (t)}} = \frac{{t^2 }}{{\sin ^2 (t)}} = \left[ {\frac{t}{{\sin (t)}}} \right]^2 \\
\end{array}\)
Does that give you any ideas?
\begin{array}{l}
\lim _{x \to 0} \left( {\frac{x}{{\sin (x)}}} \right) = \left[ {\lim _{x \to 0} \left( {\frac{{\sin (x)}}{x}} \right)} \right]^{ - 1} = 1 \\
\frac{{t^2 }}{{1 - \cos ^2 (t)}} = \frac{{t^2 }}{{\sin ^2 (t)}} = \left[ {\frac{t}{{\sin (t)}}} \right]^2 \\
\end{array}\)
Does that give you any ideas?
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Use the trig limits they gave you:
. . . . .lim<sub>t->0</sub> [1 - cos(t)]/t = 0
. . . . .lim<sub>t->0</sub> [sin(t)]/t = 1
See where that takes you.
If you get stuck, please reply showing your work and reasoning. Thank you.
Eliz.
. . . . .lim<sub>t->0</sub> [1 - cos(t)]/t = 0
. . . . .lim<sub>t->0</sub> [sin(t)]/t = 1
See where that takes you.
If you get stuck, please reply showing your work and reasoning. Thank you.
Eliz.
pka
Elite Member
- Joined
- Jan 29, 2005
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- 11,976
The limit \(\displaystyle \lim _{x \to 0} \left( {\frac{{\sin (x)}}{x}} \right) = 1\) is one of the most important and useful limits is all mathematics.
For one thing, it tells us that for all non-zero numbers t close to zero the sin(t) and t have practically the same value. Thus \(\displaystyle {\frac{{\sin (t)}}{t}}\) is practically 1 as is \(\displaystyle \left( {\frac{{\sin (t)}}{t}} \right)^n\).
Thus wouldn’t it hold for \(\displaystyle \left( {\frac{t}{{\sin (t)}}} \right)^n ?\)
For one thing, it tells us that for all non-zero numbers t close to zero the sin(t) and t have practically the same value. Thus \(\displaystyle {\frac{{\sin (t)}}{t}}\) is practically 1 as is \(\displaystyle \left( {\frac{{\sin (t)}}{t}} \right)^n\).
Thus wouldn’t it hold for \(\displaystyle \left( {\frac{t}{{\sin (t)}}} \right)^n ?\)