In Part 1 of this series, we illustrated three of the most common tactics you must know to use in order to be able to solve limit problems in Calculus:

1. Substitution: super easy when it works, and in particular when you do not get 00 as a result. If you do obtain 00, 0 divided by 0, then use one of the approaches below.
2. Factor: If possible, factor the numerator and/or the denominator. Then do some canceling.
3. Use Conjugates: If you have square roots, then multiply the numerator and the denominator by the conjugate of the square-root part. Again, you'll be able to do some canceling.

(See Part 1 for details on those.)

In this post, we're going to look at two other tactics you'll frequently need to invoke.

I. Tactic #4: Algebraic manipulation (dust off those skills)

Sometimes when you try Substitution and you obtain 00, some straightforward algebraic manipulation and canceling is enough to do the trick. Example 1 illustrates.

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Example 1.

Find limℎ→0(ℎ−5)2−25ℎ.

Solution.

As always, we first try Substitution and simply plug ℎ =0 into the expression:
limℎ→0(ℎ−5)2−25ℎ=(0−5)2−250=25−250=00
Since the limit is in the form 00, it is indeterminate—we don't yet know what is it. So we have to do some work to turn the expression into a different form that's more helpful.

Let's try the simple move of expanding the quadratic in the numerator, and then see what happens:

limℎ→0(ℎ−5)2−25ℎ=limℎ→0(ℎ2−10ℎ+25)−25ℎ=limℎ→0ℎ2−10ℎℎ=limℎ→0ℎ(ℎ−10)ℎ=limℎ→0ℎ(ℎ−10)ℎ=limℎ→0(ℎ−10)=0−10=−10✓
So expanding the binomial, along with some cancellation, let us find the limit:
limℎ→0(ℎ−5)2−25ℎ=limℎ→0(ℎ−10)=−10

That approach works because the function we started with, (ℎ−5)2−25ℎ, and the one we ended up with after expanding and simplifying, ℎ −10, are the same—except that the first function is undefined at h = 0 (since its denominator is zero there), while the second is not. We've shown this in the side-by-side graphs below. As you can see, their limits are the same as ℎ →0.

[End Example 1.]

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If you have an expression you can expand algebraically, do so.

As Example 1 illustrates, if you have an expression you can expand algebraically, do so. After you do, and then cancel some terms, you'll be able to use simple Substitution to finish the problem.

Let's consider another example where some simple algebraic manipulation is all we need.

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Example 2.

Find lim𝑥→31𝑥−13𝑥−3.

Solution.

As always, we try Substitution first, and set 𝑥 =3 in the expression to see what we get:

lim𝑥→31𝑥−13𝑥−3=13−133−3=00
Once again we obtain 00, that indeterminate result: we don't yet know what the limit is, and we have more work to do. In this case, let's use our algebra skills to put the two fractions in the numerator over the common denominator 3𝑥: lim𝑥→31𝑥−13𝑥−3=lim𝑥→333𝑥−𝑥3𝑥𝑥−3=lim𝑥→33−𝑥3𝑥𝑥−3=lim𝑥→33−𝑥(3𝑥)(𝑥−3)=lim𝑥→3−(𝑥−3)(3𝑥)(𝑥−3)=lim𝑥→3−(𝑥−3)(3𝑥)(𝑥−3)=lim𝑥→3−13𝑥=−13(3)=−19✓

[End Example 2.]

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Do whatever algebraic manipulation you can.

The upshot: If when you try Substitution you initially get 00, then do whatever algebraic manipulation you can. Just dive in: factor, multiply by the conjugate, expand the binomial, find a common denominator . . . . Once you do, you'll be able to cancel some terms. And then, finally, you'll be able to use Substitution with the new expression you've developed, and hence finish the problem easily.

II. Trigonometric limits, and two "Special Limits"

Limits that involve trig functions like sin⁡(𝑥), cos⁡(𝑥) and tan⁡(𝑥) may also require a little algebraic manipulation . . . and some basic trig facts. For instance, always remember the trig identity
sin2⁡(𝑥)+cos2⁡(𝑥)=1

Example 3 illustrates.

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Example 3.

Find lim𝑥→01−cos⁡(𝑥)sin2⁡(𝑥).

Solution.

As always, we first try Substitution:

lim𝑥→01−cos⁡(𝑥)sin2⁡(𝑥)=1−cos⁡(0)sin2⁡(0)=1−10=00

Yet again we obtain 00, that indeterminate result: we don't yet know what the limit is. We have more work to do.

Seeing the sin2⁡(𝑥) there in the denominator makes us think about the trig identity above, which we can rewrite as

sin2⁡(𝑥)=1−cos2⁡(𝑥)

Hence

lim𝑥→01−cos⁡(𝑥)sin2⁡(𝑥)=lim𝑥→01−cos⁡(𝑥)1−cos2⁡(𝑥)=lim𝑥→01−cos⁡(𝑥)(1−cos⁡(𝑥))(1+cos⁡(𝑥))=lim𝑥→01−cos⁡(𝑥)(1−cos⁡(𝑥))(1+cos⁡(𝑥))=lim𝑥→011+cos⁡(𝑥)=11+cos⁡(0)=11+1=12✓

[End Example 3.]

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As Example 3 shows, your basic trig facts, plus again some algebra (factoring, and then canceling), all let you move toward putting the expression into a new form. One where Substitution works.

Early in your study of Calculus, there are two special limits that you just have to memorize.

In addition, early in your study of Calculus, there are two special limits that you just have to memorize:

I. lim𝑥→0sin⁡(𝑥)𝑥=1II. lim𝑥→01−cos⁡(𝑥)𝑥=0

The first one, in particular, appears frequently, albeit often in a "disguised" form. The next two examples illustrate.

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Example 4.

Find lim𝑥→0sin⁡(5𝑥)𝑥.

Solution.

To be able to use the first special limit, we need what's in the argument of the sine function to match what's in the denominator. That is, since we have sin⁡(5𝑥) in the numerator, we need 5𝑥 in the denominator. So let's multiply the expression by 55:

lim𝑥→0sin⁡(5𝑥)𝑥=lim𝑥→055⋅sin⁡(5𝑥)𝑥=5⋅lim𝑥→0sin⁡(5𝑥)5𝑥[Recall lim𝑥→0sin⁡(whatever)(the same whatever)=1]=5⋅1=5✓

[End Example 4.]

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Example 5.

Find lim𝑥→0tan⁡(𝑥)𝑥.

Solution.

Once again, we first try Substitution:

lim𝑥→0tan⁡(𝑥)𝑥=tan⁡(0)0=00

And once again, the indeterminate result 00, meaning once again we have more work to do.

In this case, we know that tan⁡(𝑥) =sin⁡(𝑥)cos⁡(𝑥), so let's start there:

lim𝑥→0tan⁡(𝑥)𝑥=lim𝑥→0sin⁡(𝑥)cos⁡(𝑥)⋅1𝑥=lim𝑥→0sin⁡(𝑥)𝑥⋅1cos⁡(𝑥)[Recall lim𝑥→0sin⁡(𝑥)𝑥=1]=1⋅1cos⁡(0)[Recall cos⁡(0)=1]=1⋅1=1✓

[End Example 5.]

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The upshot: Remember your basic trig facts, and again just dive in and do whatever algebraic manipulation you have to so you can use substitution at the end. And memorize the two special limits, so you'll recognize them if they're hidden in a problem.

Of course you need to practice.

Of course reading through our discussion isn't enough. Instead, you need to practice---and make some mistakes for yourself---so that this is all routine for you when you take your exam. We have lots of problems for you to try, all with complete solutions a single click away so you can quickly check your work, or get unstuck, with no hassle.

For now, please let us know over on the Forum:

• What questions do you have?
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• How is Calculus going for you so far?