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Evaluating Trigonometric Limits Analytically

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The limit we just proved above can help us evaluate many related trigonometric limits that initially have the indeterminate form 00.

In Example 11, we use this limit to establish limθ01cosθθ=0. This limit also proves useful in later chapters.

Example 11: Evaluating an Important Trigonometric Limit

Evaluate limθ01cosθθ.

Solution

In the first step, we multiply by the conjugate so that we can use a trigonometric identity to convert the cosine in the numerator to a sine:

limθ01cosθθ=limθ01cosθθ1+cosθ1+cosθ=limθ01cos2θθ(1+cosθ)=limθ0sin2θθ(1+cosθ)=limθ0sinθθsinθ1+cosθ=(limθ0sinθθ)(limθ0sinθ1+cosθ)=102=0.

Therefore,

limθ01cosθθ=0.

Exercise 11

Evaluate limθ01cosθsinθ.

Hint

Multiply numerator and denominator by 1+cosθ.

Answer

limθ01cosθsinθ=limθ01cosθsinθ1+cosθ1+cosθ=limθ01cos2θ(sinθ)(1+cosθ)=limθ0sin2θ(sinθ)(1+cosθ)=limθ0sinθ1+cosθ=sin01+cos0=01+1=02=0

Example 12

a. Evaluate limθ05θ3+sin5θθ3.

b. Evaluate limx0tan2(7x)5x.

c. Evaluate limx5sin(x5)x225.

Solution

In all three of these examples, we notice that plugging in the target value gives us the indeterminate form 00.  That means there is likely a way to evaluate the limit, and since these limits all can be written in terms of a sine function, we attempt to rewrite the limit expression in such a way that we can use the limit we proved above: limθ0sinθθ=1.

a.  Evaluate limθ05θ3+sin5θθ3.
limθ05θ3+sin5θθ3=limθ0[5θ3θ3+sin3θθ3sin2θ]=limθ0[5+(sinθθ)3sin2θ]=limθ05+limθ0(sinθθ)3limθ0sin2θUsing the Sum and Product Rules.=5+(limθ0sinθθ)3(limθ0sinθ)2Using the Constant Rule and Power Rule.=5+1302Using limits proven above.=5

Note how we needed to use the limit rules to rewrite this limit so that the limit, limθ0sinθθ was set off by itself.  Once we have that limit set off, we know its value from what we proved earlier and we were able to evaluate the limit clearly and without issues.

 

b. Evaluate limx0tan2(7x)5x.
limx0tan2(7x)5x=limx0[sin(7x)cos(7x)tan(7x)5x]=limx0[sin(7x)7x7tan(7x)5cos(7x)]We rewrite and multiply through by 77 to obtain form sinuu with u=7x.=(limx0sin(7x)7x)75limx0tan(7x)cos(7x)We set off limx0sin(7x)7x by itself.=175tan0cos0Now each part can be evaluated.=7501=750=0

 

c. Evaluate limx5sin(x5)x225.

This example is the most interesting and requires a little more insight.  We observe that there is a sine function in the limit expression and we see that it does have the form 00, so we will seek to rewrite it so that part of the limit can be expressed in terms of limu0sinuu for some u.

limx5sin(x5)x225=limx5sin(x5)x51x+5Here we factored the denominator and see that u=x5 is a natural choice.=limx5sin(x5)x5limx51x+5

Now, is we let u=x5, then as x5,u0, so we could rewrite limx5sin(x5)x5=limu0sinuu.

Then we have:

limx5sin(x5)x225=limx5sin(x5)x5limx51x+5=(limu0sinuu)(15+5)We can evaluate the second limit by direct substitution.=1110=110

Contributors:

Example 11 and the statement of Exercise 11 is from the original OpenStax Calculus text.

The rest, including the solution to Exercise 11 is by Paul Seeburger (Monroe Community College).


Evaluating Trigonometric Limits Analytically is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

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