Video: Finding Limits Involving Trigonometric Functions

Determine lim_(π‘₯ β†’ 0) (9 βˆ’ 9 cos 7π‘₯)/(3π‘₯).

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Video Transcript

Determine the limit as π‘₯ approaches zero of nine minus nine cos of seven π‘₯ over three π‘₯.

First, we notice that we can cancel a factor of three from the top and bottom of this fraction, leaving us with this limit. Next, we notice that we can factor three out of the limit. If we attempt to do direct substitution at this point, we will see that we obtain zero over zero, which is undefined.

Let’s instead use the fact that the limit as π‘₯ approaches zero of one minus cos of π‘₯ over π‘₯ is equal to zero. We can in fact adapt this formula in order to find that the limit as π‘₯ approaches zero of one minus cos of π‘Žπ‘₯ over π‘₯ is equal to zero, where π‘Ž is just a constant. We can do this by substituting π‘Žπ‘₯ in for π‘₯ into the first rule. We obtain that the limit as π‘Žπ‘₯ goes to zero of one minus cos of π‘Žπ‘₯ over π‘Žπ‘₯ is equal to zero.

Now since π‘Ž is a constant and π‘Žπ‘₯ is going to zero, this means that π‘₯ is also going to zero. Instead of writing π‘Žπ‘₯ goes to zero, we can simply write π‘₯ goes to zero. Next, we notice we have a factor of π‘Ž in the denominator of this fraction. And so we can factorize this π‘Ž out of the limit.

Next, we can multiply both sides of the equation by π‘Ž. Since the right-hand side is zero, zero times π‘Ž gives us zero. So the right-hand side remains as zero. And so now we have found that the limit as π‘₯ goes to zero of one minus cos of π‘Žπ‘₯ over π‘₯ is zero, which is what we were trying to show. By substituting π‘Ž equals seven into this formula, we can see that our limit here is simply zero. And since three multiplied by zero is simply zero, we find that the solution here is simply zero.

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