Do You Know A Cool Way To Solve This?

Let $\mathcal{L}=\displaystyle\lim_{x \to 1} \left( \dfrac{23}{1-x^{23}}-\dfrac{11}{1-x^{11}} \right)...............(1)$

Now put $x=\dfrac{1}{y}$

When $x$ gets closer to $1$ , $y\rightarrow 1$ .

$\therefore \mathcal{L}=\displaystyle\lim_{y \to 1} \left( \dfrac{-23y^{23}}{1-y^{23}}+\dfrac{11y^{11}}{1-y^{11}} \right)$

$\therefore \mathcal{L}=\displaystyle\lim_{x \to 1} \left( \dfrac{-23x^{23}}{1-x^{23}}+\dfrac{11x^{11}}{1-x^{11}} \right)...............(2)$ (Since $x$ and $y$ are dummy variables)

Adding $(1)$ and $(2)$ we get

$2\mathcal{L}=23-11$

$\therefore \mathcal{L}=\boxed{6}$

Consider a more general form of this limit: $L=\lim_{x \to 1} \left( \frac{p}{1-x^{p}}-\frac{q}{1-x^{q}} \right)$

1) Let $x=1+t$ with $t \to 0$ , It follows that $L=\lim_{t \to 0} \left( \frac{p}{1-(1+t)^p}-\frac{q}{1-(1+t)^q}\right)$ 2) According to taylor series expansions we have $(1+t)^a=1+at+\frac{1}{2}(a-1)at^2+O(t^3)$ Use this series expansions to rewrite $(1+t)^p$ and $(1+t)^q$ neglecting $O(t^3)$ : $L=\lim_{t \to 0} \left( \frac{p}{-pt-\frac{1}{2}(p-1)pt^2}-\frac{q}{-qt-\frac{1}{2}(q-1)qt^2}\right)$ 3) Go ahead: $\begin{array}{c}\\ L&=\lim_{t \to 0} \left( \frac{1}{t}\left( \frac{1}{-1-\frac{1}{2}(p-1)t}-\frac{1}{-1-\frac{1}{2}(q-1)t}\right)\right) \\ &=\lim_{t \to 0} \left( \frac{1}{t}\left( \frac{1}{1+\frac{1}{2}(q-1)t}-\frac{1}{1+\frac{1}{2}(p-1)t}\right)\right) \\ &=\lim_{t \to 0} \left( \frac{1}{t} \frac{\frac{1}{2}(p-1)t-\frac{1}{2}(q-1)t}{\left(1+\frac{1}{2}(q-1)t\right)\left(1+\frac{1}{2}(p-1)t\right)}\right) \\ &=\lim_{t \to 0} \left( \frac{\frac{1}{2}(p-1)-\frac{1}{2}(q-1)}{\left(1+\frac{1}{2}(q-1)t\right)\left(1+\frac{1}{2}(p-1)t\right)}\right) \\ &=\frac{p-q}{2} \end{array}$

Rewrite as $\displaystyle\lim_{x\rightarrow 1} \dfrac{(23)(1-x^{11})-(11)(1-x^{23})}{(1-x^{23})(1-x^{11})}$

If you try substituting 1 in for x, you'll get 0/0. L'Hopital's rule can be applied here.

Expand. $\displaystyle\lim_{x\rightarrow 1} \dfrac{12-23x^{11}+11x^{23}}{1-x^{11}-x^{23}+x^{34}}$

Differentiate the numerator and denominator separately. $\displaystyle\lim_{x\rightarrow 1} \dfrac{-23\cdot 11 \cdot x^{10} + 23 \cdot 11 \cdot x^{22} }{-11x^{10}-23x^{22}+34x^{33}}$

Substituting 1 for x still gives 0/0, so differentiate the numerator and denominator separately, again. $\displaystyle\lim_{x\rightarrow 1} \dfrac{-23\cdot 11 \cdot 10 \cdot x^{9} + 23 \cdot 22 \cdot 11 \cdot x^{21} }{-10\cdot 11\cdot x^{9}-23\cdot 22 \cdot x^{21}+34 \cdot 33 \cdot x^{32}}$

Substituting 1 for x now gives us $\dfrac{-23\cdot 11 \cdot 10 + 23\cdot 22\cdot 11}{-10 \cdot 11 - 23 \cdot 22 + 34 \cdot 33 }$ , which simplifies to $\boxed{6}$ .

Let the limit be $L$ .

$\displaystyle L = \lim_{x \to 1} \left( \dfrac{23}{1-x^{23}} - \dfrac{11}{1-x^{11}} \right)$

$\displaystyle L = \lim_{x \to 1} \left( \dfrac{23(1-x^{11}) - 11(1-x^{23})}{(1-x^{23})(1-x^{11})} \right)$

$\displaystyle L = \lim_{x \to 1} \left( \dfrac{12 -23x^{11} + 11x^{23}}{1 + x^{34} - x^{23} - x^{11}} \right)$

This is of the form $\dfrac{0}{0}$ . So we will use L-Hopital's Rule.

$\displaystyle L = \lim_{x \to 1} \left( \dfrac{-23.11x^{10} + 11.23x^{22}}{34x^{33} - 23x^{22} - 11x^{10}} \right)$

Which is again of the form $\dfrac{0}{0}$ .

$\displaystyle L = \lim_{x \to 1} \left( \dfrac{-23.11.10x^{9} + 11.23.22x^{21}}{34.33x^{32} - 23.22x^{21} - 11.10x^{9}} \right)$

$L = \dfrac{23.11(22-10)}{11(34.3 - 23.2 - 10)} = \dfrac{23.11(12)}{11(46)} = \boxed{6}$

Simply apply l hopitals rule

I like Karthik's ingenuity, but I think there is a missing step in his argument. As it stands it could be applied to each term separately to show, for example, that

$\displaystyle\lim_{x \to 1}\left(\frac{23}{1-x^{23}}\right)=\frac{23}{2}$ which is clearly wrong!

To make his argument watertight he would need to show that his two limits really are equal as one approaches $x=1$ from above and the other one approaches from below.

My approach was to use l'Hopital's rule. Consider the more general problem

$\displaystyle\lim_{x \to 1}\left(\frac{p}{1-x^{p}}-\frac{q}{1-x^{q}}\right)=\lim_{x \to 1}\left(\frac{p(1-x^q)-q(1-x^p)}{(1-x^{p})(1-x^q)}\right)$

Since this is indeterminate when $x=1$ we can use l'Hopital's rule to get

$\displaystyle\lim_{x \to 1}\left(\frac{pq(x^{p-1}-x^{q-1})}{-qx^{q-1}-px^{p-1}+(p+q)x^{p+q-1}}\right)$

This is still indeterminate when $x=1$ so we can use l'Hopital again to get

$\displaystyle\lim_{x \to 1}\left(\frac{pq((p-1)x^{p-2}-(q-1)x^{q-2}))}{-q(q-1)x^{q-2}-p(p-1)x^{p-2}+(p+q)(p+q-1)x^{p+q-2}}\right)$

Now at last substituting $x=1$ gives the intelligible answer $\frac{p-q}{2}$

In the given problem $p=23$ and $q=11$ leading to the answer $\boxed{6}$

I must admit that this is all rather messy, but I love the way l'Hopital's rule makes use of a seeming disaster!

If we do some algebra the original expression is equal to:

$\lim_{x\to\ 1}\left( \frac{23(1-x^{11})-11(1-x^{23})}{(1-x^{23})(1-x^{11})} \right)$

= $\lim_{x\to\ 1}\left( \frac{23(1-x^{11})-11(1-x^{23})}{(1-x)^2(1+x+x^2+...+x^{22})(1+x+x^2+...+x^{10})} \right)$

Let's factor $1-x$ from the numerator and we get:

= $\lim_{x\to\ 1}\left( \frac{(1-x)(23(1+x+x^2+...+x^{10})-11(1+x+x^2+...+x^{22})}{(1-x)^2(1+x+x^2+...+x^{22})(1+x+x^2+...+x^{10})} \right)$

We can cancel $1-x$ from the top. Now if we look at the polynomial on the numerator (let's call it $p(x)$ ) we see that:

$p(1)=23(1+1+1^2+...+1^{10})-11(1+1+1^2+...+1^{22}) = 23 \cdot 11 -11 \cdot 23 =0$ then $x-1$ is a factor of $p(x)$ .

Let's say $p(x)=q(x)(x-1)$ then by long division we get that

$q(x) = -11x^{21}-11(2)x^{20}+...+(-11)(11)x^{11}+(-11)(12)x^{10} + 12(-10)x^{9} +12(-9)x^{8}+...+12(-2)x^1+12(-1)$ cool :)

Now the original limit is just:

$\lim_{x\to\ 1}\left( \frac{q(x)}{s(x)} \right) = \frac{q(1)}{s(1)}$

Here $s(x)= -(1+x+x^2+...+x^{22})(1+x+x^2+...+x^{10})$

And the result is just $6$ . :)

(I would like to know if that method can be generalized for any $p$ and $q$ as in the other solutions)

Hey guys just cross multiply and make the limit into one fn which it u put X=1, you get 0/0 form therefore use l'hospital rule twice to get a non 0/0 answer which is 6.

Tested various selected values of $x$ and noted where values headed as $x \Rightarrow 1$ and saw the difference was approaching (relatively quickly) $\boxed{6}$ .

I know that's cheating, so to speak, but it worked. 😏

Lhopital rule always works... Double differentiate both the numerator and denominator....