Differentiation Of Trigonometric Functions Homework

Differentiation Of Trigonometric Functions Homework-37
They can’t always be done, but sometimes, such as this case, they can simplify the problem.The change of variables here is to let \(\theta = 6x\) and then notice that as \(x \to 0\) we also have \(\theta \to 6\left( 0 \right) = 0\).

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See the Proof of Trig Limits section of the Extras chapter to see the proof of these two limits. Students often ask why we always use radians in a Calculus class. The proof of the formula involving sine above requires the angles to be in radians.

If the angles are in degrees the limit involving sine is not 1 and so the formulas we will derive below would also change.

All we need to do is multiply the numerator and denominator of the fraction in the denominator by 7 to get things set up to use the fact. \[\begin\mathop \limits_ \frac & = \frac\\ & = \frac\\ & = \frac\\ & = \frac\end\] This limit looks nothing like the limit in the fact, however it can be thought of as a combination of the previous two parts by doing a little rewriting.

First, we’ll split the fraction up as follows, \[\mathop \limits_ \frac = \mathop \limits_ \frac\frac\] Now, the fact wants a \(t\) in the denominator of the first and in the numerator of the second.

To see that we can use the fact on this limit let’s do a change of variables.

A change of variables is really just a renaming of portions of the problem to make something look more like something we know how to deal with.

\[\mathop \limits_ \frac = \frac\mathop \limits_ \frac = \frac\left( 1 \right) = \frac\] Now, in this case we can’t factor the 6 out of the sine so we’re stuck with it there and we’ll need to figure out a way to deal with it.

To do this problem we need to notice that in the fact the argument of the sine is the same as the denominator ( both \(\theta \)’s).

When doing a change of variables in a limit we need to change all the \(x\)’s into \(\theta \)’s and that includes the one in the limit.

Doing the change of variables on this limit gives, \[\begin\mathop \limits_ \frac & = 6\mathop \limits_ \frac\hspace\theta = 6x\ & = 6\mathop \limits_ \frac\ & = 6\left( 1 \right)\ & = 6\hspace\end\] And there we are.


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