This is the first thing you should always try: just plug the value of x into f(x). If you obtain a number (and in particular, if you don't get  $\dfrac{0}{0}$), you have your answer and are finished. In that case, these problems are completely straightforward.

Example.
Find $\displaystyle{\lim_{x \to 2}(x^3-5x + 7)}$.

Solution. \begin{align*} \lim_{x \to 2}(x^3-5x + 7) &= (2)^3 -5(2) + 7 \\ &= 8 -10 + 7 = 5 \quad \cmark \end{align*}

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Practice this (simple!) tactic in the next few problems. The solutions are immediately available using the Show/Hide Solution toggle. #5 illustrates an important point.

Frequently (on homework and exams, anyway) when you try Substitution, you obtain the fraction  $\dfrac{0}{0}$.  When this happens, if you can factor the numerator and/or the denominator, do so. A common term will cancel, removing the problematic  $0$  in the denominator. Guaranteed. Example. Find $\displaystyle{\lim_{x \to 2}\frac{x^2-4}{x-2}}$. Solution. We first try substitution: \begin{align*} \lim_{x \to 2}\frac{x^2-4}{x-2} &= \frac{2^2 -4}{2 - 2} = \frac{0}{0} \end{align*} Because this limit is in the form of  $\dfrac{0}{0}$,  it is "indeterminate"—we don't yet know what it is. So let's factor the numerator: \begin{align*} \lim_{x \to 2}\frac{x^2-4}{x-2} &= \frac{2^2 -4}{2 - 2} = \frac{0}{0} \end{align*} Because this limit is in the form of  $\dfrac{0}{0}$,  it is "indeterminate"—we don't yet know what it is. So let's factor the numerator: \begin{align*} \lim_{x \to 2}\frac{x^2-4}{x-2} &= \lim_{x \to 2}\frac{(x+2)(x-2)}{x-2} \\[8px] \text{Ah, now we can cancel the } &\text{problematic term:} &\phantom{= \lim_{x \to 2}\frac{(x+2)(x-2)}{x-2}}\\[8px] &= \lim_{x \to 2}\frac{(x+2)\cancel{(x-2)}}{\cancel{x-2}} \\[8px] &= \lim_{x \to 2} \,(x+2) \\[8px] \text{And now easy Substitution }& \text{to finish:} \\[8px] &= 2 + 2 = 4 \quad \cmark \end{align*}

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Practice the tactic of factoring to find the limit in the next few problems. These are straightforward once you learn to recognize what to do. Note: Every Calculus exam on limits that we've ever seen has at least one problem that requires this tactic.

If you try Substitution and obtain the fraction  $\dfrac{0}{0}$  and the expression has a square root in it, then rationalize the expression just like you practiced in Algebra. That is, multiply the numerator and the denominator by the conjugate of the part that has the square root. The following example illustrates.

Example. Find $\displaystyle{ \lim_{x \to 0}\dfrac{\sqrt{x+5} - \sqrt{5}}{x}}$.

Solution. As always, we first try Substitution:

$$\lim_{x \to 0}\dfrac{\sqrt{x+5} - \sqrt{5}}{x} = \frac{\sqrt{5} - \sqrt{5}}{0} = \dfrac{0}{0}$$ Since this limit is in the form of  $\dfrac{0}{0}$,  it is indeterminate—we don't yet know what it is. We have more work to do.

So let's try to rationalize the expression. The square-root part is $\sqrt{x+5} - \sqrt{5}$, so we multiply both the numerator and the denominator by the conjugate $\dfrac{\sqrt{x+5} + \sqrt{5}}{\sqrt{x+5} + \sqrt{5}}=1$:

\begin{align*} \lim_{x \to 0}\dfrac{\sqrt{x+5} - \sqrt{5}}{x} &= \lim_{x \to 0}\dfrac{\sqrt{x+5} - \sqrt{5}}{x} \cdot \dfrac{\sqrt{x+5} + \sqrt{5}}{\sqrt{x+5} + \sqrt{5}} \\[8px] &= \lim_{x \to 0}\dfrac{\sqrt{x+5}\sqrt{x+5} + \sqrt{x+5}\sqrt{5} - \sqrt{5}\sqrt{x+5} -\sqrt{5}\sqrt{5}}{x[\sqrt{x+5} + \sqrt{5}]} \\[8px] &= \lim_{x \to 0}\dfrac{\sqrt{x+5}\sqrt{x+5} + \cancel{\sqrt{x+5}\sqrt{5}} - \cancel{\sqrt{5}\sqrt{x+5}} -\sqrt{5}\sqrt{5}}{x[\sqrt{x+5} + \sqrt{5}]} \\[8px] &= \lim_{x \to 0}\dfrac{(x+5) - 5}{x[\sqrt{x+5} + \sqrt{5}]} \\[8px] &= \lim_{x \to 0}\dfrac{x}{x[\sqrt{x+5} + \sqrt{5}]} \\[8px] &= \lim_{x \to 0}\dfrac{\cancel{x}}{\cancel{x}[\sqrt{x+5} + \sqrt{5}]} \\[8px] &= \lim_{x \to 0}\dfrac{1}{\sqrt{x+5} + \sqrt{5}} \\[8px] &= \dfrac{1}{\sqrt{0+5} + \sqrt{5}} = \dfrac{1}{2\sqrt{5}} \quad \cmark \end{align*} Notice that when we multiplied by the conjugate, we multiplied out all of the terms in the numerator, because that's how we get rid of the square root. But we didn't multiply the terms in the denominator; instead we kept writing it as $x \left(\sqrt{x+5} + \sqrt{5} \right)$. That's because a few steps later the x canceled. Something similar will always happen, so in that early step don't multiply out the part that you didn't set out to rationalize. Instead just carry those terms along for a while, until you can cancel something.

For some problems, when you try Substitution and obtain  $\dfrac{0}{0}$,  you just use your algebraic skills to transform the expression into something where Substitution will work. For instance, if there's a quadratic, expand it. Or if you have some fractions in the numerator, put them over a common denominator.

The following problems illustrate.

When your problem has a trig function in it (sin, cos, tan, ...), often you need to do some manipulation in order to find the limit. To help:

1. Remember your fundamental trig identities. For instance:

Our Handy Table of Trig Formulas and Identities could be useful.

2. Early in the semester, there are two "Special Limits" 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*}

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The following problems illustrate how you'll frequently have to make use of these facts.