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 \(\displaystyle{\lim_{x\rightarrow-1}x^2}\).
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 \(\displaystyle{\lim_{x\rightarrow0}\frac{x}{\lvert x\rvert}}.\)
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 \(\displaystyle{\lim_{x\rightarrow-\infty}\left(-x^2+6x-5\right)}.\)
Rewriting the expression, we have
\[\begin{align} \lim_{x\rightarrow-\infty}\left(-x^2+6x-5\right) &=\lim_{x\rightarrow-\infty}\left(-(x-3)^2+4\right)\\ &=4-\lim_{x\rightarrow-\infty}(x-3)^2. \end{align}\]
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 \(\displaystyle{\lim_{x\rightarrow \infty}\frac{1}{x+7}}.\)
Since the denominator approaches infinity, i.e. \(x+7 \rightarrow \infty\) as \(x\rightarrow\infty,\) the answer is
\[\lim_{x\rightarrow \infty}\frac{1}{x+7}=\frac{1}{\displaystyle{\lim_{x\rightarrow \infty}(x+7)}}=0. \ _\square\]