## 17Calculus Limits - Solve By Factoring

##### 17Calculus

Evaluating Limits By Factoring

### Finite Limits, Infinite Limits, Limits At Infinity . . . Terminology Explained

The use of the terms finite limits, infinite limits and limits at infinity are used differently in various books and your instructor may have their own idea of what they mean. In this panel, we will try to break down the cases and explain the various ways these terms can be used as well as how we use them here at 17calculus.

When we talk about limits, we are looking at the $$\displaystyle{ \lim_{x \to c}{f(x)} = L }$$. The various terms apply to the description of $$c$$ and $$L$$ and are shown in the table below. The confusion lies with the terms finite limits and infinite limits. They can mean two different things.

$$\displaystyle{ \lim_{x \to c}{f(x)} = L }$$

when

term(s) used

$$c$$ is finite

limits approaching a finite value or finite limits

$$c$$ is infinite $$\pm \infty$$

limits at infinity or infinite limits

$$L$$ is finite

finite limits

$$L$$ is infinite $$\pm \infty$$

infinite limits

You can see where the confusion lies. The terms finite limits and infinite limits are used to mean two different things, referring to either $$c$$ or $$L$$. It is possible to have $$c = \infty$$ and $$L$$ be finite. So is this an infinite limit or a finite limit? It depends if you are talking about $$c$$ or $$L$$.

How 17calculus Uses These Terms
The pages on this site are constructed based on what $$c$$ is, i.e. we use the terms finite limits and infinite limits based on the value of $$c$$ only ( using the first two rows of the table above and ignoring the last two ). This seems to be the best way since, when we are given a problem, we can't tell what $$L$$ is until we finish the problem, and therefore we are unable to determine what type of problem we have and know what techniques to use until we are done with the problem.

Important: Make sure and check with your instructor to see how they use these terms.

This is the second technique for evaluating finite limits. It is very powerful and you will use it a lot in conjunction with substitution.

Topics You Need To Understand For This Page

Once we have substituted the limit value in the function and get the indeterminate form $$0/0$$, we can try factoring and canceling to get a determinate form. This technique works best with polynomials or with terms that can factor.

When we get $$0/0$$, we have a hole in the graph (a removable discontinuity). In order to understand why this works, we invoke the following theorem.

Theorem: Functions That Agree At All But One Point [proof]

Let $$c$$ be a real number and let $$g(x) = h(x)$$ for all $$x \neq c$$ in an open interval containing $$c$$.

If $$\displaystyle{ \lim_{x \to c}{[g(x)]} }$$ exists, then $$\displaystyle{ \lim_{x \to c}{[h(x)]} }$$ also exists and $$\displaystyle{ \lim_{x \to c}{[h(x)]} = \lim_{x \to c}{[g(x)]} }$$.

What this theorem says is that if we have two functions that are identical, expect (possibly) at $$x=c$$, and we know that one function has a limit that exists at $$c$$, then the other one is guaranteed to have a limit that exists at $$c$$ and that limit is the same as the limit of the first function.

This theorem is incredibly useful for functions with holes (our current topic). If we have a function with a hole, and if we can find a second function that is the same as first function, except at $$x=c$$, that has a limit there, then we can use that second function to determine the limit of the first function. Usually, this second function is simpler than the first and the way we find it is by factoring and canceling common terms.

Example

Let $$\displaystyle{ f(x) = \frac{x^2-1}{x-1} = \frac{(x-1)(x+1)}{x-1} }$$     and     $$g(x) = x+1$$

Notice that by canceling the $$(x-1)$$ term in $$f(x)$$, we get $$g(x)$$.   However, $$f(x) \neq g(x)$$ since the domains of the two functions are different. The domain of $$f(x)$$ is $$x \neq 1$$ and the domain of $$g(x)$$ is all real numbers. This can be seen in these plots. Notice there is a hole at $$x=1$$ for $$f(x)$$. This hole occurs because the factor $$x-1$$ appears in both the numerator and denominator of $$f(x)$$. $$\displaystyle{ f(x) = \frac{x^2-1}{x-1} = \frac{(x-1)(x+1)}{x-1} }$$

Domain: $$\{ x ~|~ x \in \mathbb{R} ~ \text{ and } ~ x \neq 1 \}$$ $$g(x) = x+1$$

Domain: $$\{ x ~|~ x \in \mathbb{R} \}$$

However, since $$f(x)$$ is equal to $$g(x)$$ except for the one point $$x=1$$, we can say that $$\displaystyle{ \lim_{x \to 1}{[f(x)]} = \lim_{x \to 1}{[g(x)]} }$$ by applying the above theorem. Notice the difference, the functions are not equal but the limits at $$x=1$$ are equal. This is an important distinction.

Okay, let's try factoring on some practice problems.

Practice

Use factoring to solve these limits, giving your answers in exact, simplified form.

Basic

$$\displaystyle{ \lim_{x\rightarrow 2}{\left[\frac{3x^2-x-10}{x^2-4}\right]} }$$

Problem Statement

Evaluate this limit by factoring. $$\displaystyle{\lim_{x\rightarrow 2}{\left[\frac{3x^2-x-10}{x^2-4}\right]}}$$

$$\displaystyle{\lim_{x\rightarrow 2}{\left[\frac{3x^2-x-10}{x^2-4}\right]}=11/4}$$

Problem Statement

Evaluate this limit by factoring. $$\displaystyle{\lim_{x\rightarrow 2}{\left[\frac{3x^2-x-10}{x^2-4}\right]}}$$

Solution

 First of all, you always want to try direct substitution. When you do this on this equation, you get $$0/0$$ which is indeterminate. There are at least two ways to solve this. Since this is a ratio of polynomials, we will use algebra. You can also use L'Hôpital's Rule. First, we factor the numerator and denominator and cancel like terms. $$\displaystyle{\lim_{x \rightarrow 2} \left[ \frac{3x^2-x-10}{x^2-4} \right] = }$$ $$\displaystyle{ \lim_{x \rightarrow 2} \left[ \frac{(x-2)(3x+5)}{(x-2)(x+2)} \right]}$$ Now the $$(x-2)$$ term that appears in both the numerator and denominator can cancel. $$\displaystyle{\lim_{x \rightarrow 2} \left[ \frac{3x+5}{x+2} \right]}$$ Now let's try substitution again. $$\displaystyle{\lim_{x \rightarrow 2} \left[ \frac{3x+5}{x+2} \right] = \frac{3(2)+5}{2+2} = 11/4 }$$

$$\displaystyle{\lim_{x\rightarrow 2}{\left[\frac{3x^2-x-10}{x^2-4}\right]}=11/4}$$

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$$\displaystyle{\lim_{x\rightarrow 3}{\left[\frac{x^4-81}{2x^2-5x-3}\right]}}$$

Problem Statement

Evaluate this limit by factoring. $$\displaystyle{\lim_{x\rightarrow 3}{\left[\frac{x^4-81}{2x^2-5x-3}\right]}}$$

$$\displaystyle{\lim_{x \rightarrow 3}{ \left[\frac{x^4-81}{2x^2-5x-3}\right] }= 108/7 }$$

Problem Statement

Evaluate this limit by factoring. $$\displaystyle{\lim_{x\rightarrow 3}{\left[\frac{x^4-81}{2x^2-5x-3}\right]}}$$

Solution

 First of all, you always want to try direct substitution. When you do this in this limit, you get $$0/0$$ which is indeterminate.There are at least two ways to solve this. We will work it using algebra (since this is a ratio of polynomials). (You can also use L'Hôpital's Rule, since direct substitution yields an indeterminate form.) First, we factor the numerator and denominator and cancel like terms. $$\displaystyle{\lim_{x \rightarrow 3} \left[ \frac{x^4-81}{2x^2-5x-3} \right] }$$ $$\displaystyle{ \lim_{x \rightarrow 3} \left[ \frac{(x^2-9)(x^2+9)}{(x-3)(2x+1)} \right] }$$ $$\displaystyle{\lim_{x \rightarrow 3} \left[ \frac{(x-3)(x+3)(x^2+9)}{(x-3)(2x+1)} \right]}$$ Now the $$(x-3)$$ term that appears in both the numerator and denominator can cancel. $$\displaystyle{\lim_{x \rightarrow 3} \left[ \frac{(x+3)(x^2+9)}{(2x+1)} \right] }$$ Now let's try substitution again. $$\displaystyle{ \lim_{x \rightarrow 3} \left[ \frac{(x+3)(x^2+9)}{(2x+1)} \right] = \frac{(3+3)(3^2+9)}{(2(3)+1)} = 108/7}$$

$$\displaystyle{\lim_{x \rightarrow 3}{ \left[\frac{x^4-81}{2x^2-5x-3}\right] }= 108/7 }$$

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$$\displaystyle{ \lim_{x\to2}{\frac{x^2-5x+6}{x^2-4}} }$$

Problem Statement

Evaluate this limit by factoring. $$\displaystyle{ \lim_{x\to2}{\frac{x^2-5x+6}{x^2-4}} }$$

Solution

### Krista King Math - 13 video solution

video by Krista King Math

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$$\displaystyle{ \lim_{x\to1}{\frac{x^2+x-2}{x^2-4x+3}} }$$

Problem Statement

Evaluate this limit by factoring. $$\displaystyle{ \lim_{x\to1}{\frac{x^2+x-2}{x^2-4x+3}} }$$

$$\displaystyle{\lim_{x\to1}{\frac{x^2+x-2}{x^2-4x+3}}=-3/2}$$

Problem Statement

Evaluate this limit by factoring. $$\displaystyle{ \lim_{x\to1}{\frac{x^2+x-2}{x^2-4x+3}} }$$

Solution

She made a mistake in notation here. In the last line of the first column, she wrote $$\displaystyle{ \lim_{x \to 1}{\frac{x+2}{x-3}}}$$ which is correct. However, in the first line of the next column, she should have written $$\displaystyle{\frac{1+2}{1-3}=\frac{3}{-2}}$$ WITHOUT the limit as x goes to 1 in front of it. Once you substitute the value, you do not write the limit.

### Krista King Math - 14 video solution

video by Krista King Math

$$\displaystyle{\lim_{x\to1}{\frac{x^2+x-2}{x^2-4x+3}}=-3/2}$$

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$$\displaystyle{ \lim_{x\to-1}{\frac{x+1}{x^2-x-2}} }$$

Problem Statement

Evaluate this limit by factoring. $$\displaystyle{ \lim_{x\to-1}{\frac{x+1}{x^2-x-2}} }$$

Solution

### Krista King Math - 15 video solution

video by Krista King Math

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$$\displaystyle{ \lim_{t\to-3}{\frac{t^2+6t+9}{t^2-9}} }$$

Problem Statement

Evaluate this limit by factoring. $$\displaystyle{ \lim_{t\to-3}{\frac{t^2+6t+9}{t^2-9}} }$$

Solution

### Krista King Math - 16 video solution

video by Krista King Math

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$$\displaystyle{ \lim_{z\to-2}{\frac{(z+2)^2}{z^4-16}} }$$

Problem Statement

Evaluate this limit by factoring. $$\displaystyle{ \lim_{z\to-2}{\frac{(z+2)^2}{z^4-16}} }$$

Solution

### Krista King Math - 17 video solution

video by Krista King Math

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$$\displaystyle{ \lim_{x\to2}{\frac{x^2-4}{x-2}} }$$

Problem Statement

Evaluate this limit by factoring. $$\displaystyle{ \lim_{x\to2}{\frac{x^2-4}{x-2}} }$$

Solution

### PatrickJMT - 25 video solution

video by PatrickJMT

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$$\displaystyle{\lim_{h\to0}{\frac{(4+h)^2-16}{h}}}$$

Problem Statement

Evaluate this limit by factoring. $$\displaystyle{\lim_{h\to0}{\frac{(4+h)^2-16}{h}}}$$

Solution

### PatrickJMT - 26 video solution

video by PatrickJMT

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$$\displaystyle{ \lim_{x\to-1}{ \frac{2x+2}{x+1} } }$$

Problem Statement

Evaluate the limit $$\displaystyle{ \lim_{x\to-1}{ \frac{2x+2}{x+1} } }$$ by factoring.

$$\displaystyle{ \lim_{x\to-1}{ \frac{2x+2}{x+1} } }$$ $$= 2$$

Problem Statement

Evaluate the limit $$\displaystyle{ \lim_{x\to-1}{ \frac{2x+2}{x+1} } }$$ by factoring.

Solution

 $$\displaystyle{ \lim_{x\to-1}{ \frac{2x+2}{x+1} } }$$ $$\displaystyle{ \lim_{x\to-1}{ \frac{2(x+1)}{x+1} } }$$ The $$x+1$$ terms in the numerator and denominator cancel. $$\displaystyle{ \lim_{x\to-1}{ 2 } = 2 }$$

$$\displaystyle{ \lim_{x\to-1}{ \frac{2x+2}{x+1} } }$$ $$= 2$$

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$$\displaystyle{\lim_{x\to3}{\frac{x^2-6x+9}{x^2-9}}}$$

Problem Statement

Evaluate this limit by factoring. $$\displaystyle{\lim_{x\to3}{\frac{x^2-6x+9}{x^2-9}}}$$

Solution

### Khan Academy - 33 video solution

video by Khan Academy

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$$\displaystyle{\lim_{x\to1}{\frac{x^2+x-2}{x-1}}}$$

Problem Statement

Evaluate this limit by factoring. $$\displaystyle{\lim_{x\to1}{\frac{x^2+x-2}{x-1}}}$$

Solution

### Khan Academy - 34 video solution

video by Khan Academy

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$$\displaystyle{ \lim_{z\to2}{ \frac{z^2+2z-8}{z^4-16} } }$$

Problem Statement

Evaluate the limit $$\displaystyle{ \lim_{z\to2}{ \frac{z^2+2z-8}{z^4-16} } }$$ by factoring, giving your answer in exact terms.

$$\displaystyle{ \lim_{z\to2}{ \frac{z^2+2z-8}{z^4-16} } }$$ $$= 3/16$$

Problem Statement

Evaluate the limit $$\displaystyle{ \lim_{z\to2}{ \frac{z^2+2z-8}{z^4-16} } }$$ by factoring, giving your answer in exact terms.

Solution

 $$\displaystyle{ \lim_{z\to2}{ \frac{z^2+2z-8}{z^4-16} } }$$ $$\displaystyle{ \lim_{z\to2}{ \frac{(z+4)(z-2)}{(z^2+4)(z^2-4)} } }$$ $$\displaystyle{ \lim_{z\to2}{ \frac{(z+4)(z-2)}{(z^2+4)(z+2)(z-2)} } }$$ $$\displaystyle{ \lim_{z\to2}{ \frac{z+4}{(z^2+4)(z+2)} } }$$ $$\displaystyle{ \frac{2+4}{(2^2+4)(2+2)} }$$ $$\displaystyle{ = \frac{6}{(8)(4)} }$$ $$\displaystyle{ = \frac{3}{(8)(2)} }$$ $$\displaystyle{ = \frac{3}{16} }$$

To check my answer, I decided to plot the rational function. $$\displaystyle{ \lim_{z\to2}{ \frac{z^2+2z-8}{z^4-16} } }$$ $$= 3/16$$

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Intermediate

$$\displaystyle{ \lim_{t \rightarrow -2}{ \left[ \frac{\frac{1}{t}+\frac{1}{2}}{t^3+8}\right]} }$$

Problem Statement

Evaluate this limit by factoring. $$\displaystyle{ \lim_{t \rightarrow -2}{ \left[ \frac{\frac{1}{t}+\frac{1}{2}}{t^3+8}\right]} }$$

$$\displaystyle{\lim_{t\rightarrow -2}{\left[\frac{\frac{1}{t}+\frac{1}{2}}{t^3+8}\right]}=-1/48}$$

Problem Statement

Evaluate this limit by factoring. $$\displaystyle{ \lim_{t \rightarrow -2}{ \left[ \frac{\frac{1}{t}+\frac{1}{2}}{t^3+8}\right]} }$$

Solution

The first thing you want to try here is to plug in -2 for t. However, in this limit, we get 0/0, which is an indeterminate form. So let's try factoring. But first we need to get the fraction in a different form.
In the numerator, we have $$\displaystyle{ \frac{1}{t} + \frac{1}{2} = \frac{2}{2t} + \frac{t}{2t} = \frac{t+2}{2t} }$$.

So the limit looks like $$\displaystyle{\lim_{t \rightarrow -2}{ \left[ \frac{ \frac{t+2}{2t}}{t^3+8} \right]} = \lim_{t \rightarrow -2}{ \left[ \frac{t+2}{(2t)(t^3+8)} \right]} }$$.

Now we need to factor the denominator. Let's check quickly to make sure (t+2) is a factor. This is a great time to practice synthetic division. Using synthetic division, $$t^3+8$$ factors into $$(t+2)(t^2-2t+4)$$.
So our limit is now $$\displaystyle{\lim_{t \rightarrow -2}{ \left[ \frac{t+2}{(2t)(t+2)(t^2-2t+4)} \right]} }$$. We can cancel the (t+2) term in the numerator and denominator and get $$\displaystyle{\lim_{t \rightarrow -2} \left[ \frac{1}{(2t)(t^2-2t+4)} \right]}$$.
Substituting -2 for t gives us $$\displaystyle{ \frac{1}{(-4)((-2)^2-2(-2)+4)} = }$$ $$\displaystyle{ \frac{1}{-4(4+4+4)} = -1/48 }$$

$$\displaystyle{\lim_{t\rightarrow -2}{\left[\frac{\frac{1}{t}+\frac{1}{2}}{t^3+8}\right]}=-1/48}$$

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$$\displaystyle{ \lim_{x\rightarrow 0}{\left[\frac{x^3-7x}{x^3}\right]} }$$

Problem Statement

Evaluate this limit by factoring. $$\displaystyle{ \lim_{x\rightarrow 0}{\left[\frac{x^3-7x}{x^3}\right]} }$$

$$\displaystyle{\lim_{x\rightarrow 0}{\left[\frac{x^3-7x}{x^3}\right]}=-\infty}$$

Problem Statement

Evaluate this limit by factoring. $$\displaystyle{ \lim_{x\rightarrow 0}{\left[\frac{x^3-7x}{x^3}\right]} }$$

Solution

$$\displaystyle{\lim_{x\rightarrow 0}{\left[\frac{x^3-7x}{x^3}\right]} = }$$ $$\displaystyle{ \lim_{x\rightarrow 0}{\left[ \frac{x(x^2-7)}{x(x^2)} \right]} = }$$ $$\displaystyle{ \lim_{x \rightarrow 0}{\left[ \frac{x^2-7}{x^2} \right]} }$$
Direct substitution yields $$-7/0$$. So, the trick here is determine if the limit is $$+\infty$$, $$-\infty$$ or DNE (does not exist). To do this, we need to look at the limit as $$x$$ approaches zero from the left and compare it to the limit from the right. If they are different, then the limit does not exist.
Let's look at the left side of zero first. When x is less than zero (negative), $$x^2$$ is always positive. So the numerator is negative and the denominator is positive telling us that $$\displaystyle{ \lim_{x \rightarrow 0^- }{\left[ \frac{x^2-7}{x^2} \right]}= -\infty }$$.
Now, looking at the right side of zero, x is positive and $$x^2$$ is also positive. Since the numerator is negative and the denominator is positive, we know that $$\displaystyle{ \lim_{x \rightarrow 0^+ }{\left[ \frac{x^2-7}{x^2} \right]}= -\infty }$$.
Since $$\displaystyle{ \lim_{x \rightarrow 0^- }{\left[ \frac{x^2-7}{x^2} \right]}= \lim_{x \rightarrow 0^- }{\left[ \frac{x^2-7}{x^2} \right]}= -\infty }$$, the limit $$\displaystyle{ \lim_{x \rightarrow 0 }{\left[ \frac{x^2-7}{x^2} \right]}= -\infty }$$

$$\displaystyle{\lim_{x\rightarrow 0}{\left[\frac{x^3-7x}{x^3}\right]}=-\infty}$$

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$$\displaystyle{ \lim_{z\rightarrow 1}{\left[\frac{z^3-1}{(z-1)^2}\right]} }$$

Problem Statement

Evaluate this limit by factoring. $$\displaystyle{ \lim_{z\rightarrow 1}{\left[\frac{z^3-1}{(z-1)^2}\right]} }$$

$$\displaystyle{ \lim_{z\rightarrow 1}{ \left[ \frac{z^3-1}{(z-1)^2}\right] } }$$ does not exist.

Problem Statement

Evaluate this limit by factoring. $$\displaystyle{ \lim_{z\rightarrow 1}{\left[\frac{z^3-1}{(z-1)^2}\right]} }$$

Solution

Direct substitution yields $$0/0$$ which is indeterminate.
Factoring yields
$$\displaystyle{ \lim_{z \rightarrow 1}{\left[\frac{z^3-1}{(z-1)^2}\right]} = }$$ $$\displaystyle{\lim_{z \rightarrow 1}{\left[\frac{(z-1)(z^2+z+1)}{(z-1)^2}\right]} = }$$ $$\displaystyle{\lim_{z \rightarrow 1}{ \left[\frac{z^2+z+1}{z-1}\right]}}$$
Direct substitution here gives us $$3/0$$. So is the answer $$+\infty$$, $$-\infty$$ or DNE (without looking at a graph)?
Well, considering values close to 1, when $$z < 1$$ the result is negative and when $$z>1$$ the result is positive. So the answer is neither $$+\infty$$ nor $$-\infty$$ and the only correct is answer is that the limit does not exist.

$$\displaystyle{ \lim_{z\rightarrow 1}{ \left[ \frac{z^3-1}{(z-1)^2}\right] } }$$ does not exist.

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$$\displaystyle{ \lim_{x\to1}{ \frac{x^3-1}{x^4-1} } }$$

Problem Statement

Evaluate this limit by factoring. $$\displaystyle{ \lim_{x\to1}{ \frac{x^3-1}{x^4-1} } }$$

Solution

### Krista King Math - 18 video solution

video by Krista King Math

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$$\displaystyle{ \lim_{h\to0}{\frac{(3+h)^{-1}-3^{-1}}{h}} }$$

Problem Statement

Evaluate this limit by factoring. $$\displaystyle{ \lim_{h\to0}{\frac{(3+h)^{-1}-3^{-1}}{h}} }$$

Solution

### PatrickJMT - 27 video solution

video by PatrickJMT

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$$\displaystyle{\lim_{x\rightarrow 27}{\left[\frac{x-27}{x^{1/3}-3}\right]}}$$

Problem Statement

Evaluate this limit by factoring. $$\displaystyle{\lim_{x\rightarrow 27}{\left[\frac{x-27}{x^{1/3}-3}\right]}}$$

$$\displaystyle{ \lim_{x \rightarrow 27}{\left[\frac{x-27}{x^{1/3}-3}\right]}=27}$$

Problem Statement

Evaluate this limit by factoring. $$\displaystyle{\lim_{x\rightarrow 27}{\left[\frac{x-27}{x^{1/3}-3}\right]}}$$

Solution

 Of course, the first thing we want to try is to plug in 27 directly for x. This gives us $$0/0$$ which is indeterminate. So we need to do something else.Upon initial inspection, this expression doesn't seem to be factorable. However, I am going to show you a trick. I don't know about you but I am not completely comfortable working with fractional powers. So I am going to substitute $$t$$ for $$x^{1/3}$$. We can do that as long we do the substitution for every expression containing x. So what do we get? $$\displaystyle{ \frac{x-27}{x^{1/3}-3} = \frac{t^3-27}{t-3} }$$ Notice that since $$t=x^{1/3}$$, we can cube both sides to get $$t^3=x$$.Using synthetic division, we can factor the numerator which becomes $$(t^3-27)=(t-3)(t^2+3t+9)$$.Now the expression is $$\displaystyle{ \frac{(t-3)(t^2+3t+9)}{t-3}}$$ and we can easily see that the $$t-3$$ will cancel.So let's go back to x's since our limit is in terms of x. $$\displaystyle{ \lim_{x \rightarrow 27}{\frac{x-27}{x^{1/3}-3}}}$$ $$\displaystyle{ \lim_{x \rightarrow 27}{\frac{(x^{1/3}-3)(x^{2/3}+3x^{1/3}+9)}{x^{1/3}-3}} }$$ $$\displaystyle{ \lim_{x \rightarrow 27}{\frac{(x^{2/3}+3x^{1/3}+9)}{1}} }$$ $$27^{2/3}+3(27)^{1/3}+9 = 27$$ Note - - We could have used L'Hôpital's Rule here, which would have been much easier.

$$\displaystyle{ \lim_{x \rightarrow 27}{\left[\frac{x-27}{x^{1/3}-3}\right]}=27}$$

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$$\displaystyle{\lim_{x\rightarrow 1}{\left[\frac{x^{1/3}-1}{x^{1/4}-1}\right]}}$$

Problem Statement

Evaluate this limit by factoring. $$\displaystyle{\lim_{x\rightarrow 1}{\left[\frac{x^{1/3}-1}{x^{1/4}-1}\right]}}$$

$$\displaystyle{\lim_{x\rightarrow 1}{\left[\frac{x^{1/3}-1}{x^{1/4}-1}\right]}=4/3}$$

Problem Statement

Evaluate this limit by factoring. $$\displaystyle{\lim_{x\rightarrow 1}{\left[\frac{x^{1/3}-1}{x^{1/4}-1}\right]}}$$

Solution

 Direct substitution yields $$0/0$$, which is indeterminate. To solve this exercise we need to do a trick. We want to see if we can factor this problem to cancel terms in the numerator and denominator. I don't know about you but I have a hard time working with fractional powers. So I am going to use substitution to remove fractional powers. In this problem, I am going to substitute for x. I want to get rid of both the 1/3rd and 1/4th powers. So, to do that I am going to let $$t = x^{ 1/12 }$$. I chose 1/12 since 1/12=(1/3)(1/4). You will see in a minute why this works. Okay, so we have $$t=x^{1 / 12}$$ which means $$t^{12}=x$$ (by taking the 12th power of both sides). Now, let's substitute. $$\displaystyle{ \frac{x^{1/3}-1}{x^{1/4}-1} = \frac{(t^{12})^{1/3}-1}{(t^{12})^{1/4}-1} = \frac{t^4-1}{t^3-1} }$$ Okay, so I can factor the numerator and denominator here (use synthetic division) and I get $$\displaystyle{ \frac{(t-1)(t^3+t^2+t+1)}{(t-1)(t^2+t+1)}}$$ Even though this is not completely factored, we can see that we can cancel $$(t-1)$$ in this fraction, so we can stop factoring here.Okay, let's go back to limit notation and convert back to x's. $$\displaystyle{ \lim_{x \rightarrow 1}{\frac{(x^{1/12}-1)(x^{1/4}+x^{1/6}+x^{1/12}+1)}{(x^{1/12}-1)(x^{1/6}+x^{1/12}+1)}} }$$ $$\displaystyle{\lim_{x \rightarrow 1}{\frac{x^{1/4}+x^{1/6}+x^{1/12}+1}{x^{1/6}+x^{1/12}+1}}}$$ Now when we substitute 1 for x, we get $$\displaystyle{ \frac{1+1+1+1}{1+1+1} = 4/3 }$$ Note - - We could have used L'Hôpital's Rule here, which would have been much easier.

$$\displaystyle{\lim_{x\rightarrow 1}{\left[\frac{x^{1/3}-1}{x^{1/4}-1}\right]}=4/3}$$

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