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Evaluate the Limit lim t→0 of (tan(9t))/(sin(3t)):
a) 3
b) 9
c) 1/3
d) 0

User Kaksat
by
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1 Answer

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Final answer:

To evaluate the limit lim t→0 of (tan(9t))/(sin(3t)), we can use L'Hospital's Rule. Applying the rule, we take the derivatives of the numerator and denominator, and evaluate the limit again. The limit is 3.

Step-by-step explanation:

To evaluate the limit lim t→0 of (tan(9t))/(sin(3t)), we can use L'Hospital's Rule which states that if we have an indeterminate form 0/0 or ∞/∞, we can find the limit by taking the derivative of the numerator and denominator and then evaluating the limit again. Let's apply L'Hospital's Rule:

  1. Take the derivative of tan(9t) with respect to t: d/dt[tan(9t)] = 9sec²(9t)
  2. Take the derivative of sin(3t) with respect to t: d/dt[sin(3t)] = 3cos(3t)
  3. Now evaluate the limit again: lim t→0 of (9sec²(9t))/(3cos(3t))
  4. Plug t = 0 into the derivatives: (9sec²(0))/(3cos(0)) = 9/(3*1) = 3

Therefore, the limit lim t→0 of (tan(9t))/(sin(3t)) is 3.

User Matthew Hood
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