Video Transcript
Find the limit as 𝑥 approaches 𝜋 by two of five sin 𝑥 minus five over two 𝑥 minus 𝜋.
First, we can try solving this limit by using direct substitution. This gives us five sin of 𝜋 by two minus five over two times 𝜋 by two minus 𝜋. Since sin of 𝜋 by two is equal to one, this simplifies to five minus five over 𝜋 minus 𝜋, which is equal to zero over zero and is therefore undefined.
However, it does enable us to use L’Hôpital’s rule, which tells us that if the limit as 𝑥 approaches some limit 𝑎 of 𝑓 of 𝑥 over 𝑔 of 𝑥 is equal to zero over zero, positive infinity over positive infinity, or negative infinity over negative infinity. And if 𝑓 of 𝑥 and 𝑔 of 𝑥 are both differentiable near 𝑎, then the limit as 𝑥 approaches 𝑎 of 𝑓 of 𝑥 over 𝑔 of 𝑥 is equal to the limit as 𝑥 approaches 𝑎 of 𝑓 prime of 𝑥 over 𝑔 prime of 𝑥.
Now, in our case, 𝑓 of 𝑥 will be five sin of 𝑥 minus five. And 𝑔 of 𝑥 will be two 𝑥 minus 𝜋. Since our limit is equal to zero over zero and both 𝑓 and 𝑔 are differentiable, we’re able to apply L’Hôpital’s rule. We’re required to find 𝑓 prime of 𝑥 and 𝑔 prime of 𝑥. Differentiating five sin of 𝑥 minus five with respect to 𝑥 gives us that 𝑓 prime of 𝑥 is equal to five cos 𝑥. And differentiating two 𝑥 minus 𝜋 with respect to 𝑥 gives us that 𝑔 prime of 𝑥 must be equal to two.
Now we can apply L’Hôpital’s rule to say that our limit must be equal to the limit as 𝑥 approaches 𝜋 by two of five cos over two. And we can apply direct substitution to this limit to say that it’s equal to five cos of 𝜋 by two over two. Cos of by 𝜋 two is equal to zero.
Therefore, we can conclude that our limit must be equal to zero.
