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Question
Let f(t) = `|(cos"t","t", 1),(2sin"t", "t", 2"t"),(sin"t", "t", "t")|`, then `lim_("t" - 0) ("f"("t"))/"t"^2` is equal to ______.
Options
0
– 1
2
3
Solution
Let f(t) = `|(cos"t","t", 1),(2sin"t", "t", 2"t"),(sin"t", "t", "t")|`, then `lim_("t" - 0) ("f"("t"))/"t"^2` is equal to 0.
Explanation:
We have f(t) = `|(cos"t","t", 1),(2sin"t", "t", 2"t"),(sin"t", "t", "t")|`
Expanding along R1
= `cos "t"|("t", 2"t"),("t", "t")| - "t"|(2 sin "t", 2"t"),(sin"t", "t")| + 1|(2 sin "t", "t"),(sin"t", "t")|`
= cos t(t2 – 2t) – t(2t sint – 2t sin t) + (2t sin t – t sin t)
= –t2 cos t + t sin t
∴ `("f"("t"))/"t"^2 = ("t"^2 cos"t" + "t" sin"t")/"t"^2`
⇒ `("f"("t"))/"t"^2 = - cos "t" + (sin "t")/"t"`
⇒ `lim_("t" -> 0) ("f"("t"))/"t"^2 = lim_("t" -> 0) (- cos "t") + lim_("t" -> 0) (sin "t")/"t"`
= – 1 + 1
= 0
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