Video: APCALC02AB-P1A-Q01-738102145093

Find lim_(π‘₯β†’πœ‹/2) (5sin π‘₯ βˆ’ 5) / (2π‘₯ βˆ’ πœ‹).

01:46

Video Transcript

Find the limit as π‘₯ approaches πœ‹ by two of five sin π‘₯ minus five over two π‘₯ minus πœ‹.

First, we can try solving this limit by using direct substitution. This gives us five sin of πœ‹ by two minus five over two times πœ‹ by two minus πœ‹. Since sin of πœ‹ by two is equal to one, this simplifies to five minus five over πœ‹ minus πœ‹, which is equal to zero over zero and is therefore undefined.

However, it does enable us to use L’HΓ΄pital’s rule, which tells us that if the limit as π‘₯ approaches some limit π‘Ž of 𝑓 of π‘₯ over 𝑔 of π‘₯ is equal to zero over zero, positive infinity over positive infinity, or negative infinity over negative infinity. And if 𝑓 of π‘₯ and 𝑔 of π‘₯ are both differentiable near π‘Ž, then the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ over 𝑔 of π‘₯ is equal to the limit as π‘₯ approaches π‘Ž of 𝑓 prime of π‘₯ over 𝑔 prime of π‘₯.

Now, in our case, 𝑓 of π‘₯ will be five sin of π‘₯ minus five. And 𝑔 of π‘₯ will be two π‘₯ minus πœ‹. Since our limit is equal to zero over zero and both 𝑓 and 𝑔 are differentiable, we’re able to apply L’HΓ΄pital’s rule. We’re required to find 𝑓 prime of π‘₯ and 𝑔 prime of π‘₯. Differentiating five sin of π‘₯ minus five with respect to π‘₯ gives us that 𝑓 prime of π‘₯ is equal to five cos π‘₯. And differentiating two π‘₯ minus πœ‹ with respect to π‘₯ gives us that 𝑔 prime of π‘₯ must be equal to two.

Now we can apply L’HΓ΄pital’s rule to say that our limit must be equal to the limit as π‘₯ approaches πœ‹ 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œ‹ by two over two. Cos of by πœ‹ two is equal to zero.

Therefore, we can conclude that our limit must be equal to zero.

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