Math  /  Calculus

QuestionUse L'Hopital to determine the following limit (where tt is regarded as a constant): limω50cos(ωt)cos(50t)2500ω2\lim _{\omega \rightarrow 50} \frac{\cos (\omega t)-\cos (50 t)}{2500-\omega^{2}}

Studdy Solution

STEP 1

What is this asking? This problem is asking us to find the value that the expression approaches as ω\omega gets really close to 50, using L'Hôpital's rule because we get a tricky "zero divided by zero" situation if we just plug in 50 for ω\omega directly. Watch out! Remember, L'Hôpital's rule only works when we have an indeterminate form like 0/0 or infinity/infinity.
Also, don't forget to differentiate with respect to ω\omega, not tt! tt is a constant in this problem.

STEP 2

1. Check if L'Hôpital's rule applies
2. Apply L'Hôpital's rule
3. Evaluate the limit

STEP 3

When ω\omega approaches **50**, the numerator becomes cos(50t)cos(50t)=0\cos(50t) - \cos(50t) = \textbf{0}.

STEP 4

When ω\omega approaches **50**, the denominator becomes 2500502=25002500=02500 - 50^2 = 2500 - 2500 = \textbf{0}.

STEP 5

Since we have the indeterminate form 0/0\textbf{0/0}, we can **apply** L'Hôpital's rule!

STEP 6

The derivative of cos(ωt)\cos(\omega t) with respect to ω\omega is tsin(ωt)-t\sin(\omega t).
The derivative of cos(50t)-\cos(50t) with respect to ω\omega is **0** since we treat tt as a constant.
So, the derivative of the numerator is tsin(ωt)-t\sin(\omega t).

STEP 7

The derivative of 25002500 with respect to ω\omega is **0**.
The derivative of ω2-\omega^2 with respect to ω\omega is 2ω-2\omega.
So, the derivative of the denominator is 2ω-2\omega.

STEP 8

Our new limit is: limω50tsin(ωt)2ω \lim_{\omega \rightarrow 50} \frac{-t\sin(\omega t)}{-2\omega}

STEP 9

Now we can substitute ω=50\omega = 50 into the expression: tsin(50t)250 \frac{-t\sin(50t)}{-2 \cdot 50}

STEP 10

We can simplify this to: tsin(50t)100=tsin(50t)100 \frac{-t\sin(50t)}{-100} = \frac{t\sin(50t)}{100}

STEP 11

The limit is tsin(50t)100\frac{t\sin(50t)}{100}.

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