Example 12
a. Evaluate limθ→05θ3+sin5θθ3.
b. Evaluate limx→0tan2(7x)5x.
c. Evaluate limx→5sin(x−5)x2−25.
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: limθ→0sinθθ=1.
a. Evaluate limθ→05θ3+sin5θθ3.
limθ→05θ3+sin5θθ3=limθ→0[5θ3θ3+sin3θθ3⋅sin2θ]=limθ→0[5+(sinθθ)3⋅sin2θ]=limθ→05+limθ→0(sinθθ)3⋅limθ→0sin2θUsing the Sum and Product Rules.=5+(limθ→0sinθθ)3⋅(limθ→0sinθ)2Using the Constant Rule and Power Rule.=5+13⋅02Using limits proven above.=5
Note how we needed to use the limit rules to rewrite this limit so that the limit, limθ→0sinθθ 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 limx→0tan2(7x)5x.
limx→0tan2(7x)5x=limx→0[sin(7x)cos(7x)⋅tan(7x)5x]=limx→0[sin(7x)7x⋅7tan(7x)5cos(7x)]We rewrite and multiply through by 77 to obtain form sinuu with u=7x.=(limx→0sin(7x)7x)⋅75limx→0tan(7x)cos(7x)We set off limx→0sin(7x)7x by itself.=1⋅75⋅tan0cos0Now each part can be evaluated.=75⋅01=75⋅0=0
c. Evaluate limx→5sin(x−5)x2−25.
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 limu→0sinuu for some u.
limx→5sin(x−5)x2−25=limx→5sin(x−5)x−5⋅1x+5Here we factored the denominator and see that u=x−5 is a natural choice.=limx→5sin(x−5)x−5⋅limx→51x+5
Now, is we let u=x−5, then as x→5,u→0, so we could rewrite limx→5sin(x−5)x−5=limu→0sinuu.
Then we have:
limx→5sin(x−5)x2−25=limx→5sin(x−5)x−5⋅limx→51x+5=(limu→0sinuu)⋅(15+5)We can evaluate the second limit by direct substitution.=1⋅110=110