Finding Limits of Specific Functions

Find

\[lim_{x \rightarrow 2} \dfrac{2x^2-3x+1}{x^3+4}\]

Answer:

The idea is to apply the Quotient Rule for limits if possible. Since the numerator and denominator are both polynomials.

\[lim_{x \rightarrow 2} (2x^2-3x+1)=2(2)^2-3(2+1\]

\[lim_{x \rightarrow 2} = 8-6+1\]

\[lim_{x \rightarrow 2}=3\]

and

\[lim_{x \rightarrow 2}(x^3+4)=2^3+4\]

\[lim_{x \rightarrow 2}(x^3+4)=12\]

which means the conditions to apply the Quotient Rule for limits is met. Now you know that:

\[lim_{x \rightarrow 2} \dfrac{2x^2-3x+1}{x^3+4}=\dfrac{3}{12}\]

\[lim_{x \rightarrow 2} = \dfrac{1}{4}\].

Now find

\[\lim_{x \rightarrow \infty} \dfrac{x^2+2x+4}{x^3-8}\].

Answer:

While the Quotient Rule holds for limits at infinity, it does require that the limit of the numerator and denominator are both real numbers, which in this case isn't true. That means you can't apply the Quotient Rule for limits at infinity. Instead, try factoring to see if that will help. If you factor the denominator, you will see that

\[x^3-8=(x-2)(x^2+2x+4)\]

Let's cancel some factors out! This leaves us with

\[lim_{x \rightarrow \infty } \dfrac{x^2+2x+4}{x^3-8}=lim_{x \rightarrow \infty} \dfrac{x^2+2x+4}{(x-2)(x^2+2x+4)}\]

\[lim_{x \rightarrow \infty } \dfrac{1}{x-2}\]

This is a much simpler limit; for more information on limits like this, see Infinite Limits. There you will learn how to show that

\[lim_{x \rightarrow \infty} \dfrac{1}{x-2}=0\].

That means

\[lim_{x \rightarrow \infty} \dfrac{x^2+2x+4}{x^3-8}=0\]

Find

\[lim_{ x \rightarrow 2} \dfrac{x^2+2x+4}{x^3-8}\].

Answer:

In the previous example, you were able to factor the denominator, enabling you to look at a simpler limit:

\[lim_{x \rightarrow 2} \dfrac{1}{x-2}\]

For this example, you will need to look at the limit from the left and the limit from the right and see if they are the same. In fact

\[lim_{x \rightarrow 2^+} \dfrac{1}{x-2}=\infty\]

while

\[lim_{x \rightarrow 2^-} \dfrac{1}{x-2}=-\infty\]

So, you can't find the limit, and you would say that the limit doesn't exist.

Find

\[\lim_{x \rightarrow 2} \dfrac{\dfrac{1}{x+1}-\dfrac{1}{3}}{x-2}\]

Answer:

Notice that this function has something interesting going on at \(x=2\) since the denominator is zero there. It will either be a hole in the graph, or a vertical asymptote, or some other discontinuity. That means you can't apply the Quotient Rule for limits since the limit of the denominator cannot be zero. Instead, let's do some algebra first:

\[\dfrac{\dfrac{1}{x+1}-\dfrac{1}{3}}{x-2} = \left( \dfrac{1}{x-2} \right) \left( \dfrac{1}{x+1}-\dfrac{1}{3} \right)\]

\[\left( \dfrac{1}{x-2} \right) \left( \dfrac{3-(x+1)}{3(x+1)} \right)\]

\[\left( \dfrac{1}{x-2} \right) \left( \dfrac{2-x}{3(x+1)} \right)\]

\[-\left( \dfrac{1}{x-2} \right) \left( \dfrac{x-2}{3(x+1)} \right)\]

\[-\dfrac{1}{3} \left( \dfrac{1}{x+1} \right)\]

which means that

\[lim_{x \rightarrow 2} \dfrac{\dfrac{1}{x+1}-\dfrac{1}{3}}{x-2}=-\dfrac{1}{9}\]

Find

\[\lim_{x \rightarrow 2} \dfrac{\sqrt{x+3}-1}{x+2}\]

Answer:

Again you are blocked from using the Quotient Rule for limits because the limit of the denominator is zero if you plug in -2. So try multiplying both the numerator and denominator by the conjugate of the numerator:

\[\dfrac{\sqrt{x+3}-1}{x+2}=\dfrac{(\sqrt{x+3}-1)(\sqrt{x+3}+1)}{(x+2)(\sqrt{x+3}+1)}\]

\[\dfrac{(\sqrt{x+3})^2-1}{(x+2)(\sqrt{x+3}+1)}\]

\[\dfrac{x+3-1}{(x+2)(\sqrt{x+3}+1)}\]

\[\dfrac{x+2}{(x+2)(\sqrt{x+3}+1)}\]

\[\dfrac{1}{\sqrt{x+3}+1}\]

Now try evaluating the limit of the denominator, and you will see that

\[lim_{x \rightarrow -2} (\sqrt{x+3}+1)=\sqrt{-2+3}+1\]

\[\sqrt{1}+1\]

.

This means you can apply the Quotient Rule for limits to say that

\[lim_{x \rightarrow -2} \dfrac{1}{\sqrt{x+3}+1}=\dfrac{1}{\sqrt{1}+1}\].

Now you know that

\[lim_{x \rightarrow -2} \dfrac{\sqrt{x+3}-1}{x+2}=\dfrac{1}{2}\].

Using the function

\[f(x)= \left( \begin{matrix} x+3 &, x \geq 1\\ x^2-4 &, x<1\end{matrix} \right) \]

find

\[lim_{x \rightarrow 1} f(x)\]

if it exists.

Answer:

If this were any limit other than as \(x \rightarrow 1\) you could plug in function values to find the limit since both pieces of the function are polynomials. But \(x=1\) is where the definition of the function changes, so instead you need to look at the limit from the left and the limit from the right. For this function.

\[lim_{x \rightarrow 1^+} f(x)= lim_{x \rightarrow 1^+}(x+3)=4 \]

and

\[lim_{x \rightarrow 1^-} f(x)=lim_{x \rightarrow 1^-}(x^2-4)=-3\]

Since those two numbers aren't the same

\[lim_{x \rightarrow 1} f(x)\]

does not exist.

Find

\[lim_{x \rightarrow 2} e^{\sqrt{x-1}}\].

Answer:

Think of this limit as the composition of two functions,

\(f(x)=e^x\) and \(g(x)=\sqrt{x-1}\)

Then

\[f \cdot g(x)=f(g(x))=e^{\sqrt{x-1}} \]

You already know that the exponential function is continuous everywhere, and that \(g(x)\) has a limit as \(x \rightarrow 2\). Therefore

\[lim_{x \rightarrow 2} f(g(x))=f \left(lim_{x \rightarrow 2} g(x) \right)\]

\[f\left( lim_{x \rightarrow 2} \sqrt{x-1} \right)\]

\[f(\sqrt{2-1})\]

\[f(1)=e^1=e\]