### Video Transcript

Find the limit as π₯ approaches π by two of five sin π₯ minus five over two π₯ 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 LβHΓ΄pitalβs rule, which tells us that if the limit as π₯ approaches some limit π of π of π₯ over π of π₯ is equal to zero over zero, positive infinity over positive infinity, or negative infinity over negative infinity. And if π of π₯ and π of π₯ are both differentiable near π, then the limit as π₯ approaches π of π of π₯ over π of π₯ is equal to the limit as π₯ approaches π of π prime of π₯ over π prime of π₯.

Now, in our case, π of π₯ will be five sin of π₯ minus five. And π of π₯ will be two π₯ minus π. Since our limit is equal to zero over zero and both π and π are differentiable, weβre able to apply LβHΓ΄pitalβs rule. Weβre required to find π prime of π₯ and π prime of π₯. Differentiating five sin of π₯ minus five with respect to π₯ gives us that π prime of π₯ is equal to five cos π₯. And differentiating two π₯ minus π with respect to π₯ gives us that π prime of π₯ must be equal to two.

Now we can apply LβHΓ΄pitalβs rule to say that our limit must be equal to the limit as π₯ approaches π by two of five cos over two. And we can apply direct substitution to this limit to say that itβs 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.