Limit with a Zing

Getting a common denominator gives us $\frac{x^a-3x+2}{x^{a+1}-x^a-2x^2+3x-1}$ Using L'Hopital's rule gives us $\frac{ax^{a-1}-3}{(a+1)x^a-ax^{a-1}-4x+3}$ We can see that this goes to infinity except when $a=3$ in which case it is a 0/0 situation. Plugging in $a=3$ and using L'Hoptal's rule again gives us $\frac{6x}{12x^2-6x-4}$ Therefore both a and the limit are 3.

From the factorization:

$x^a-2x+1= (x-1)(x^{a-1}+x^{a-2}+\cdots + x - 1),$

we have

$\displaystyle\frac{1}{x-1} - \frac{1}{x^a-2x+1} = \frac{x^{a-1}+x^{a-2}+\cdots+x-2}{x^a-2x+1}.$

Since the limit of the above expression exists and $\lim_{x\to 1} (x^a-2x+1) = 0$ , it follows that $\lim_{x\to 1} (x^{a-1}+x^{a-2}+\cdots+x-2) = 0,$ that is $(a-1)-2 = 0$ yielding $a=3.$ We verify that the limit indeed exists when $a=3$ :

$\displaystyle \frac{1}{x-1} - \frac{1}{x^3-2x+1} = \frac{x^2+x-2}{(x-1)(x^2+x-1)} = \frac{(x-1)(x+2)}{(x-1)(x^2+x-1)} = \frac{x+2}{x^2+x-1}, \quad x\neq 1.$

Hence $\displaystyle\lim_{x\to 1}\left(\frac{1}{x-1} - \frac{1}{x^3-2x+1}\right) =\lim_{x\to 1}\frac{x+2}{x^2+x-1} = \frac{1+2}{1+1-1} = 3.$

We obtain $a = 3$ and $b=3$ . Therefore $a+b = 3+3 = \boxed{6}$ .

Combining the fractions gives (x^a -3x+2)/((x-1)(x^a-2x+1). If a limit exists the top must clearly tend to zero, so we require that x^a=3x-2. Replacing x by 1+t gives (1+t)^a=1+3t, so a=ln(1+3t)/ln(1+t) = (3t - 4.5t^2 + ...)/(t - t^2/2 + ...), which tends to 3 as t tends to zero. Using this value of a, the expression boils down to (x+2)/(x^2+x-1), which equals 3 when x=1. 3+3=6.

If b exists,it is easy to know (x^a-2x+1)/(x-1)=1 when lim(x→1),using L'Hopital's rule gives us a=3,then b=3

By cross multiplication of two fractions.... Limit (x^a-3x+3)÷{(x-1)(x^a-2x+1)} X->1 Here when x->1 tha value of expression gives indeterminant form i.e. 0/0. So using L'hospital rule Limit (a x^(a-1) - 3)÷{(x-1)(a x ^(a-1) -2) +(x^a-2x+1) X->1

Here if a=3 in numerator gives factors (x-1)(x+1) And denominator has (x-1) factor ...if we cancelled out these factor from numerator and denominator gives expression

Limit (3(x+1)÷{(3x^2-2) +x^2+x-1)} X->1 On solving limit wo got 3 as ans

There fore for a=3 we got ans b=3 a+b=3+3=6

First observe that $(x-1)$ is a factor of the polynomial $(x^a-2x+1)$ as seen upon substituting $x=1$ into $x^a-2x+1= 1^a-2(1)+1=1-2+1=0$ . Now we factorise $(x^a-2x+1)$ as follows:

$x^a-2x+1=x^a-x^{a-1}+x^{a-1}-x^{a-2}+...+x^2-x-x+1=(x-1)(x^{a-1}+x^{a-2}+...+x^2+x-1)$

Then $\lim_{x \to 1} \left( \dfrac{1}{x-1}-\dfrac{1}{x^a-2x+1} \right)=\lim_{x \to 1} \left( \dfrac{1}{x-1}-\dfrac{1}{(x-1)(x^{a-1}+x^{a-2}+...+x^2+x-1)} \right)$

$=\lim_{x \to 1} \left(\dfrac{(x^{a-1}+x^{a-2}+...+x^2+x-1)-1}{(x-1)(x^{a-1}+x^{a-2}+...+x^2+x-1)}\right)=\lim_{x \to 1} \left(\dfrac{x^{a-1}+x^{a-2}+...+x^2+x-2}{(x-1)(x^{a-1}+x^{a-2}+...+x^2+x-1)}\right)$

Now clearly the denominator tends to $0$ as $x$ tends to $1$ . For the limit to exist, the numerator must also tend to $0$ . As $x$ tends to $1$ , the numerator tends to $\lim_{x \to 1} (x^{a-1}+x^{a-2}+...+x^2+x-2)=(a-1)-2$ . Then:

$(a-1)-2=0$

$a-1=2$

$a=2+1$

$a=3$

Now $b=\lim_{x \to 1} \left( \dfrac{x^2+x-2}{(x-1)(x^2+x-1)} \right)= \lim_{x \to 1} \left( \dfrac{(x+2)\cancel {(x-1)}}{\cancel{(x-1)}(x^2+x-1)} \right)=\lim_{x \to 1} \left( \dfrac{x+2}{x^2+x-1} \right)=\dfrac{1+2}{1+1-1}=3$ which means $a+b=3+3=\boxed{6}$ .

Suck your mom

Wolfram Alpha

Tried different values of a until a non infinite result.

wrote this python code:

def lim(x, a): return(1/(x-1) - 1/(x *a - 2 x +1))

x = [1.1, 1.01, 1.001, 1.0001, 1.00001, 1.000001] a = [1, 2, 3, 4, 5]

print("a\tx\tlim") print("==========================") for a now in a: for x now in x: print(a now, "\t", x now, "\t", lim(x now, a now))

In the output you can see that for a=3 the limit is 3. Obviously this is a proof for nothing, but I'm amazed every time again how easy it is to verify complicated things easily... :-)

note: the code might not be displayed well since it's not a code editor, but try to figure it out...

Using long division algorithm we can see that $(x^a-2x+1)$ is always divisible by $(x-1)$ resulting in

$A=x^{a-1}+x^{a-2}+\dots+x-1=\left(\sum_{n=1}^{a-1}{x^{n}} \right)-1$

For the limit to have a solution we need to get rid of that bottom factor of $(x-1)$ , so the hope is that after expanding the left fraction with $A$

$\frac{A}{(x-1)A}-\frac{1}{(x-1)A}$

and adding these two fractions, now with a common denominator, together

$\frac{A-1}{(x-1)A}$

the top will be divisible by $x-1$ .

Using the long division again: $(A-1)/(x-1)=(x^{a-1}+x^{a-2}+\dots+x-2)/(x-1)=x^{a-2}+2x^{a-3}+\dots+(a-2)x+(a-1)+\frac{a-3}{x-1}=\sum_{n=0}^{a-2}(a-1-n)x^n + \frac{a-3}{x-1}$

As we said the top must be divisible, so the reminder $\frac{a-3}{x-1}$ must be zero, implying a=3.

Plugging it back we get $a+b=3+\lim_{x\to1}\left(\frac{(x-1)(x-2)}{(a-1)(x^3+x^2+x-1)}\right)=3+\lim_{x\to1}\left(\frac{x+2}{x^2+x-1}\right)=3+3=6$

This sad fact about this problem is that this solution is easily brute-forceable, taking away the joy of solving it properly. However, if you want to give this to someone else, there is an easy fix: increasing the numerator of the right fraction by $n$ also increases $a$ by $n$ .