Find limit for rational function

[late347]

Homework Statement

f
[/B]
f(x)= x^2 +4
find the limit as x approaches 1, there is something wrong with the latex code but I don't know what.
Limit $$\lim_{x\to 1} \frac{{f(x)}^4-{f(1)}^4}{x-1}$$

Homework Equations

-methods for finding limits
-factorising polynomials
-possibly polynomial long division (I used it)

The Attempt at a Solution

[/B]
Well, I was a little bit bored today and decided to try to do some "more difficult" problems in my textbook for limits. You are free to comment on my polynomial long division if you want to. I tried to use it for this problem. Because for rational function if you can cancel out the pesky denominator... then you can plug in the x= something and get the limit that way.The value of the rational clause comes out at 0/0 just something to note in the beginning. Existence of limit as x->1 is still possible. We cannot immediately plug in x=1 and get the value of the limit of the rational function.

But, we ought to try to cancel the clause of function f so we are able to plug in x=1

If the expression is not cancellable, then we are going to have to start tabulating values for plus-side limit and minus-side limit, and evaluate if the both-sided limit exists.

I did the problem the tedious and long way, but it seemed to bring the correct answer eventually.

I was wondering if there's any easier way (faster way, smarter way) to get the factorization correct. Well... you could put the expression into wolfram alpha and have the computer factor it.

Foiling out the expression with a little bit pen-and-paper calculation

$[(x^4+8x^2+16)^2 -625]/(x-1)$

=
foiling out more with pen-and-paper
$(x^8+16x^6+96x^4+256x^2-369)/(x-1)$

putting that into polynomial long division in the Anglo-American format of the long division.

After some time, the remainder comes out at zero (which is good) because it implies our original expression was indeed factorable.

pic of polynomial long division. I attempted to follow the instruction in the Finnish language wiki article as best as I could. Anglo_american long division is indeed taught as the only long division in most schools I reckon...

The quotient is $(x^7+x^6+17x^5+17x^4+113x^3+113x^2+369x+369)$
and remainder =0

Hence we can continue and cancel the expression with (x-1) out of there$\lim_{x\to 1}\frac{[x-1](x^7+x^6+17x^5+17x^4+113x^3+113x^2+369x+369)}{x-1}$

$\lim_{x\to 1}{(x^7+x^6+17x^5+17x^4+113x^3+113x^2+369x+369)}$

plug in x=1 and we will have the limit = 1000, as x->1

 

[fresh_42]

Well, the result seems to be correct. But why haven't you defined $g(x)=(x^2+4)^4$ instead? If you write the limit with $g(x)$ instead of $f(x)$, does it look familiar to you?

 

[late347]

Well, the result seems to be correct. But why haven't you defined $g(x)=(x^2+4)^4$ instead? If you write the limit with $g(x)$ instead of $f(x)$, does it look familiar to you?

no I'm sort of re-learning this hopefully properly this time...

 

[fresh_42]

no I'm sort of re-learning this hopefully properly this time...

Do you know, how the differential at a point is defined?

 

[late347]

Do you know, how the differential at a point is defined?

it's in the next chapter in my textbook but yes, I have a basic idea how it would be done. https://simple.wikipedia.org/wiki/Difference_quotient

in our college last year, we did thing such as graphic derivation or whatever it was called. So I'm a little bit familiar with it. Maybe I will read more into the matter with my textbook today and tomorrow.

 

[fresh_42]

You might have a look into this

https://en.wikipedia.org/wiki/Derivative#Rigorous_definition

instead. It shortens the calculation essentially. Maybe the exercise has been intended to prepare the next chapter. So, yes, $1,000$ is the solution (and long division always worth practicing ).

 

[Mark44]

there is something wrong with the latex code but I don't know what.

You had \lim_{x\to\1}. You shouldn't have \1 (i.e., with a backslash in front of 1), and you shouldn't cram everything in so tightly, with no spaces between things.

 

[Ray Vickson]

Homework Statement

f
[/B]
f(x)= x^2 +4
find the limit as x approaches 1, there is something wrong with the latex code but I don't know what.
Limit $$\lim_{x\to 1} \;r(x) = \frac{{f(x)}^4-{f(1)}^4}{x-1}$$

Since you already have the solution, I feel OK about offering a simpler solution that avoids long division and almost everything else. First a fairly standard trick: since your ratio $r(x)$ has $x^2$ in the numerator, it is convenient to try to get $x^2$ in the denominator; we can do that by multiplying and dividing by $x+1$, to get
$$r(x) = (x+1) \frac{(x^2+4)^4 - 5^4}{x^2-1}.$$
To make this even simpler, let $x^2-1 = z$, so
$$r(x) = (x+1) \frac{(z+5)^4 - 5^4}{z},$$
and we are looking at the $z \to 0$ limit. Since the numerator vanishes when $z = 0$, it will be a polynomial of degree 4 in $z$, and with no constant term; that is, $(5+z)^4 - 5^4 = c_4 z^4 + c_3 z^3 + c_2 z^2 + c_1 z = z(c_1 + c_2 z + c_3 z^2 + c_4 z^3).$ Therefore, the ratio is just $r(x) = (x+1) (c_1 + c_2 z + c_3 z^2 + c_4 z^3)$, whose limit as $x \to 1$ (and $z \to 0$) is just $2 c_1$. Here, $c_1$ is the coefficient of $z^1$ in the expansion of $(5+z)^4.$

 

[late347]

Since you already have the solution, I feel OK about offering a simpler solution that avoids long division and almost everything else. First a fairly standard trick: since your ratio $r(x)$ has $x^2$ in the numerator, it is convenient to try to get $x^2$ in the denominator; we can do that by multiplying and dividing by $x+1$, to get
$$r(x) = (x+1) \frac{(x^2+4)^4 - 5^4}{x^2-1}.$$
To make this even simpler, let $x^2-1 = z$, so
$$r(x) = (x+1) \frac{(z+5)^4 - 5^4}{z},$$
and we are looking at the $z \to 0$ limit. Since the numerator vanishes when $z = 0$, it will be a polynomial of degree 4 in $z$, and with no constant term; that is, $(5+z)^4 - 5^4 = c_4 z^4 + c_3 z^3 + c_2 z^2 + c_1 z = z(c_1 + c_2 z + c_3 z^2 + c_4 z^3).$ Therefore, the ratio is just $r(x) = (x+1) (c_1 + c_2 z + c_3 z^2 + c_4 z^3)$, whose limit as $x \to 1$ (and $z \to 0$) is just $2 c_1$. Here, $c_1$ is the coefficient of $z^1$ in the expansion of $(5+z)^4.$

Where do you get the whole business about $c_1,~c_2...$#

Can you get some of those from pascal's triangle. I got lost a little bit there at the end also I got lost a little bit about the end conclusion that limit should be 2*$c_1$

I think it could be easier to see what's going on if I saw it in limit notation... maybe I'll try to go thru the motions of that calculation with pen and paper first since I am writing on my phone now

 

[Ray Vickson]

Where do you get the whole business about $c_1,~c_2...$#

Can you get some of those from pascal's triangle. I got lost a little bit there at the end also I got lost a little bit about the end conclusion that limit should be 2*$c_1$

I think it could be easier to see what's going on if I saw it in limit notation... maybe I'll try to go thru the motions of that calculation with pen and paper first since I am writing on my phone now

The quantity $(z+5)^4-5^4$ is a polynomial of degree 4 in $z$, so the powers $z^1, z^2, z^3, z^4$ have some numerical coefficients---call them four different names, like $c_1, c_2, c_3, c_4$. They are easy enough to get directly if we do some more work: $(z+5)^4 - 5^4 = z^4+20 z^3+150 z^2+500 z$, but why bother to calculate them until we need them? It turns out that we do not need to know that $c_4 = 1, c_3 = 20, c_2 = 150$; in the end, all we need is $c_1 = 500$. Why is that enough? Well, what do YOU think is the answer to $\lim_{z \to 0} (c_4 z^3 + c_3 z^2 + c_2 z + c_1)?$

 

[late347]

The quantity $(z+5)^4-5^4$ is a polynomial of degree 4 in $z$, so the powers $z^1, z^2, z^3, z^4$ have some numerical coefficients---call them four different names, like $c_1, c_2, c_3, c_4$. They are easy enough to get directly if we do some more work: $(z+5)^4 - 5^4 = z^4+20 z^3+150 z^2+500 z$, but why bother to calculate them until we need them? It turns out that we do not need to know that $c_4 = 1, c_3 = 20, c_2 = 150$; in the end, all we need is $c_1 = 500$. Why is that enough? Well, what do YOU think is the answer to $\lim_{z \to 0} (c_4 z^3 + c_3 z^2 + c_2 z + c_1)?$

well, for an expression like that, it looks like it is continuous, so you could just plug in 0 and get the value of the limit.
In this case $c_1$ is independant from z itself, so the $c_1$ remains the same. But the multiplications with z*c cancel out.
So the limit is $c_1$ in that case

But in the earlier example you actually had z in divisor, and the dividend both. So you said the z approaches zero, yes indeed it approches it... but the numerator z=0 and z the numerator vanishes? I was a little worried there

 

[Ray Vickson]

well, for an expression like that, it looks like it is continuous, so you could just plug in 0 and get the value of the limit.
In this case $c_1$ is independant from z itself, so the $c_1$ remains the same. But the multiplications with z*c cancel out.
So the limit is $c_1$ in that case

But in the earlier example you actually had z in divisor, and the dividend both. So you said the z approaches zero, yes indeed it approches it... but the numerator z=0 and z the numerator vanishes? I was a little worried there

No, I did not put $z = 0$ until after cancelling it. I never, ever, divided by 0. Basically,
$$\lim_{z \to 0} \frac{500z}{z} = \lim_{z \to 0} 500 = 500,$$
because as long as $z \neq 0$ we have $500 z / z = 500.$

 

[ecastro]

You might want to use this relationship to factor the numerator:

$\begin{equation*} x^2 - y^2 = \left(x + y\right) \cdot \left(x - y\right) \end{equation*}$

 

[late347]

No, I did not put $z = 0$ until after cancelling it. I never, ever, divided by 0. Basically,
$$\lim_{z \to 0} \frac{500z}{z} = \lim_{z \to 0} 500 = 500,$$
because as long as $z \neq 0$ we have $500 z / z = 500.$

I see that now... I had to walk through your solution with pen-and-paper. I used to think my English was rather good but sometimes it is hard to find the mathematical meaning from written text together with the math notation.

So at some point you actually canceled the Zs from divisor and dividend?

Let's try some latex code here once again, I did the similar thing in pen-paper style, but let's try the the code and see if it come out ok

begin by expanding by (x-1), we are actually there already and we did the $(a+b)(a-b)=a^2-b^2$
$$ \lim_{x\rightarrow 1} \frac{(x+1)*(x^2+4)^4-625}{x^2-1} $$$( x^2 -1 ) = z$ plug it into the divisor and inside the ()^4 term

$\begin{align} & \lim_{x \rightarrow 1;z \rightarrow 0}= (x+1)* \frac{(z+5)^4-5^4}{z} \nonumber \\ \leftrightarrow & \lim_{x \rightarrow 1; z \rightarrow 0}= (x+1) * \frac{z^4 +20z^3 +150z^2 +500z +625 -625}{z} \nonumber \\ \leftrightarrow & \lim_{x \rightarrow 1; z \rightarrow 0}= \frac{(x+1)*[z(z^3 +20z^2 +150z +500)]}{z} \nonumber \\ \end{align}$

x^2-1=z
then one z variable can be canceled from divisor and dividend

$\leftrightarrow \lim_{x \rightarrow 1}= [x+1]*[(x^2-1)^3 +20(x^2-1)^2 +150(x^2-1) +500] \nonumber \\$
looks like the limit will be sometihng like

2*(500) because now the entire polynomial thingy is continuous, because the divisor was cancellable and canceled out. We just plug in the x->1 x=1 and find the value of the limit.

 

[Ray Vickson]

I see that now... I had to walk through your solution with pen-and-paper. I used to think my English was rather good but sometimes it is hard to find the mathematical meaning from written text together with the math notation.

So at some point you actually canceled the Zs from divisor and dividend?

Let's try some latex code here once again, I did the similar thing in pen-paper style, but let's try the the code and see if it come out ok

begin by expanding by (x-1), we are actually there already and we did the $(a+b)(a-b)=a^2-b^2$
$$ \lim_{x\rightarrow 1} \frac{(x+1)*(x^2+4)^4-625}{x^2-1} $$$( x^2 -1 ) = z$ plug it into the divisor and inside the ()^4 term

$\begin{align} & \lim_{x \rightarrow 1;z \rightarrow 0}= (x+1)* \frac{(z+5)^4-5^4}{z} \nonumber \\ \leftrightarrow & \lim_{x \rightarrow 1; z \rightarrow 0}= (x+1) * \frac{z^4 +20z^3 +150z^2 +500z +625 -625}{z} \nonumber \\ \leftrightarrow & \lim_{x \rightarrow 1; z \rightarrow 0}= \frac{(x+1)*[z(z^3 +20z^2 +150z +500)]}{z} \nonumber \\ \end{align}$

x^2-1=z
then one z variable can be canceled from divisor and dividend

$\leftrightarrow \lim_{x \rightarrow 1}= [x+1]*[(x^2-1)^3 +20(x^2-1)^2 +150(x^2-1) +500] \nonumber \\$
looks like the limit will be sometihng like

2*(500) because now the entire polynomial thingy is continuous, because the divisor was cancellable and canceled out. We just plug in the x->1 x=1 and find the value of the limit.

Even shorter: use $a^4 - b^4 = (a-b)(a^3 + a^2 b + a b^2 + b^3),$ so
$$\begin{array}{rcl}(x^2+4)^4-5^4 &=& (x^2+4-5)((x^2+4)^3 +5 (x^2+4)^2 + 5^2 (x^2+4) + 5^3) \\
&=& (x-1)(x+1) ((x^2+4)^3 +5 (x^2+4)^2 + 5^2 (x^2+4) + 5^3)
\end{array}$$
We can cancel out the $(x-1)$ factor, then put $x=1$ in the rest of the terms to get the limiting ratio.