Limits by Substitution

A limit is defined as the value a function approaches as the variable within that function gets nearer and nearer to a particular value. Suppose we have a limit described as ${\lim_{x\rightarrow a}f(x)}$. This indicates the value of $f(x)$ when $x$ is infinitely close to $a,$ but not exactly equal to $a$. The substitution rule is a method of finding limits, by simply substituting $x$ with $a$. The mathematical manifestation of this rule would be

$$\displaystyle{\lim_{x\rightarrow a}f(x)=f(a)}.$$

Let's try out a few examples first.

Find the value of $\displaystyle{\lim_{x\rightarrow-1}x^2}$.

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This is simple. Just use the substitution rule and plug in $x=-1$, and we have

$$\displaystyle{\lim_{x\rightarrow-1}x^2}=\left.x^2\right|_{x=-1}=(-1)^2=1.\ _\square$$

However, the substitution rule does not always hold. In order to use the substitution rule, the function $f(x)$ must satisfy the following condition:

$$\text{"The function }f(x)\text{ must be continuous."}$$

This means that the graph of $f(x)$ does not break up anywhere within its domain. An example of a discontinuous function is $f(x)=\frac{1}{x}$. Try drawing this. You will notice that the graph breaks up at $x=0$, and thus it is discontinuous at $x=0$, so we cannot use the substitution rule when finding $\displaystyle{\lim_{x\rightarrow0}\frac{1}{x}}$. In fact, this limit does not exist at all, but we will discuss this later on.

Find the value of $\displaystyle{\lim_{x\rightarrow0}\frac{x}{\lvert x\rvert}}.$

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discontinuous

The figure depicts the graph of the function $f(x)=\frac{x}{\lvert x\rvert}.$ Observe that the graph is discontinuous at $x=0$, which means that we cannot apply the substitution rule to find the given limit. $_\square$

So, now our discussion comes to a simple and explicit conclusion: "If the function is continuous, just substitute the variable with the value it converges to!"

Find $\displaystyle{\lim_{x\rightarrow-\infty}\left(-x^2+6x-5\right)}.$

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Rewriting the expression, we have

$$\begin{aligned} \lim_{x\rightarrow-\infty}\left(-x^2+6x-5\right) &=\lim_{x\rightarrow-\infty}\left(-(x-3)^2+4\right)\\ &=4-\lim_{x\rightarrow-\infty}(x-3)^2. \end{aligned}$$

Since $(x-3)^2 \rightarrow \infty$ as $x\rightarrow-\infty,$ the answer is

$$\lim_{x\rightarrow-\infty}\left(-x^2+6x-5\right)=-\infty. \ _\square$$

Find $\displaystyle{\lim_{x\rightarrow \infty}\frac{1}{x+7}}.$

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Since the denominator approaches infinity, i.e. $x+7 \rightarrow \infty$ as $x\rightarrow\infty,$ the answer is

$$\lim_{x\rightarrow \infty}\frac{1}{x+7}=\frac{1}{\displaystyle{\lim_{x\rightarrow \infty}(x+7)}}=0. \ _\square$$