## 0 Divided by 0: Solve Limit Problems in Calculus, Part 2

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:

**Substitution**: super easy when it works, and in particular when you do*not*get $\dfrac{0}{0}$ as a result. If you*do*obtain $\dfrac{0}{0}$, 0 divided by 0, then use one of the approaches below.**Factor**: If possible, factor the numerator and/or the denominator. Then do some canceling.**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.

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

Sometimes when you try Substitution and you obtain $\dfrac{0}{0}$ , some straightforward algebraic manipulation and canceling is enough to do the trick. Example 1 illustrates.

**Example 1**.

Find $\displaystyle{\lim_{h \to 0}\dfrac{(h-5)^2 – 25}{h}}$.

*Solution*.

As always, we first try Substitution and simply plug $h=0$ into the expression:

$$\lim_{h \to 0}\dfrac{(h-5)^2 – 25}{h} = \dfrac{(0-5)^2 -25}{0} = \dfrac{25-25}{0} = \dfrac{0}{0}$$

Since the limit is in the form $\dfrac{0}{0},$ 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:

\begin{align*}

\lim_{h \to 0}\dfrac{(h-5)^2 – 25}{h} &= \lim_{h \to 0}\dfrac{(h^2 -10h + 25) – 25}{h} \\[12px]
&= \lim_{h \to 0}\dfrac{h^2 – 10h}{h} \\[12px]
&= \lim_{h \to 0}\dfrac{h(h – 10)}{h} \\[12px]
&= \lim_{h \to 0}\dfrac{\cancel{h}(h-10)}{\cancel{h}} \\[12px]
&= \lim_{h \to 0}(h – 10) \\[12px]
&= 0 – 10 = -10 \quad \cmark

\end{align*}

So expanding the binomial, along with some cancellation, let us find the limit:

$$\lim_{h \to 0}\dfrac{(h-5)^2 – 25}{h} = \lim_{h \to 0}(h – 10) = -10$$

That approach works because the function we started with, $\dfrac{(h-5)^2 – 25}{h}$, 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 $h \to 0$.

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

**Example 2**.

Find $\displaystyle{\lim_{x \to 3}\dfrac{\frac{1}{x}-\frac{1}{3}}{x-3}}$.

*Solution*.

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

$$\lim_{x \to 3}\dfrac{\frac{1}{x}-\frac{1}{3}}{x-3} = \dfrac{\frac{1}{3}-\frac{1}{3}}{3-3} = \dfrac{0}{0}$$

Once again we obtain $\dfrac{0}{0}$, 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 $3x$:

\begin{align*}

\lim_{x \to 3}\dfrac{\frac{1}{x}-\frac{1}{3}}{x-3} &= \lim_{x \to 3}\dfrac{\frac{3}{3x}-\frac{x}{3x}}{x-3} \\[12px]
&= \lim_{x \to 3}\dfrac{\frac{3-x}{3x}}{x-3} \\[12px]
&= \lim_{x \to 3}\dfrac{3-x}{(3x)(x-3)} \\[12px]
&= \lim_{x \to 3}\dfrac{-(x-3)}{(3x)(x-3)} \\[12px]
&= \lim_{x \to 3}\dfrac{-\cancel{(x-3)}}{(3x)\cancel{(x-3)}} \\[12px]
&= \lim_{x \to 3}\frac{-1}{3x}\\[12px]
&= \frac{-1}{3(3)} = -\frac{1}{9} \quad \cmark

\end{align*}

[End Example 2.]

**The upshot**: If when you try Substitution you initially get $\dfrac{0}{0}$, 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

$$\sin^2(x) + \cos^2(x) = 1$$

Example 3 illustrates.

**Example 3**.

Find $\displaystyle{\lim_{x \to 0}\dfrac{1- \cos(x) }{\sin^2(x)} }$.

*Solution*.

As always, we first try Substitution:

$$\lim_{x \to 0}\dfrac{1- \cos(x)}{\sin^2(x)} = \frac{1- \cos(0)}{\sin^2(0)} = \frac{1-1}{0} = \frac{0}{0}$$

Yet again we obtain $\dfrac{0}{0}$, that indeterminate result: we don’t yet know what the limit is. We have more work to do.

Seeing the $\sin^2(x)$ there in the denominator makes us think about the trig identity above, which we can rewrite as

$$ \sin^2(x) = 1 – \cos^2(x) $$

Hence

\begin{align*}

\lim_{x \to 0}\frac{1- \cos(x)}{\sin^2(x)} &= \lim_{x \to 0}\frac{1- \cos(x)}{1 – \cos^2(x)} \\[12px]
&= \lim_{x \to 0}\frac{1- \cos(x)}{(1 – \cos(x))(1 + \cos(x))} \\[12px]
&= \lim_{x \to 0}\frac{\cancel{1- \cos(x)}}{\cancel{(1 – \cos(x))}(1 + \cos(x))} \\[12px]
&= \lim_{x \to 0} \frac{1}{1 + \cos(x)} \\[12px]
&= \frac{1}{1 + \cos(0)} \\[12px]
&= \frac{1}{1+1} = \frac{1}{2} \quad \cmark

\end{align*}

[End Example 3.]

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.

**special limits**that you just have to memorize:

\begin{align*}

\text{I. } & \lim_{x \to 0}\frac{\sin(x)}{x} = 1 \\[16px] \text{II. } & \lim_{x \to 0}\frac{1-\cos(x)}{x} = 0

\end{align*}

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

**Example 4**.

Find $\displaystyle{\lim_{x \to 0}\dfrac{\sin(5x)}{x} }$.

*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 $\dfrac{5}{5}$:

\begin{align*}

\lim_{x \to 0}\dfrac{\sin(5x)}{x} &= \lim_{x \to 0} \frac{5}{5} \cdot \dfrac{\sin(5x)}{x} \\[8px]
&= 5 \cdot \lim_{x \to 0} \frac{\sin(5x)}{5x} &\text{[Recall $\displaystyle{\lim_{x \to 0}\dfrac{\sin(\text{whatever})}{(\text{the same whatever})} =1 }$]}\\[8px]
&= 5 \cdot 1 = 5 \quad \cmark

\end{align*}

[End Example 4.]

**Example 5**.

Find $\displaystyle{\lim_{x \to 0}\dfrac{\tan(x)}{x} }$.

*Solution*.

Once again, we first try Substitution:

$$ \lim_{x \to 0}\dfrac{\tan(x)}{x} = \frac{\tan(0)}{0} = \frac{0}{0}$$

And once again, the indeterminate result $\dfrac{0}{0}$, meaning once again we have more work to do.

In this case, we know that $\tan(x) = \dfrac{\sin(x)}{\cos(x)}$, so let’s start there:

\begin{align*}

\lim_{x \to 0}\dfrac{\tan(x)}{x} &= \lim_{x \to 0}\dfrac{\sin(x)}{\cos(x)} \cdot \frac{1}{x} \\[12px]
&= \lim_{x \to 0} \frac{\sin(x)}{x}\cdot \frac{1}{\cos(x)} &&\text{[Recall $\displaystyle{\lim_{x \to 0}\dfrac{\sin(x)}{x} =1 }$]}\\[8px]
&= 1 \cdot \frac{1}{\cos(0)} &&\text{[Recall $\cos(0) = 1$]} \\[8px]
&= 1 \cdot 1 = 1 \quad \cmark

\end{align*}

[End Example 5.]

**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.

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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.

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