Limit of a Difference of Rational Functions

[arpon]

Homework Statement

$$\lim _{x \rightarrow 1} (\frac{23}{1-x^{23}}-\frac{11}{1-x^{11}})$$

Homework Equations

i) For functions f and g which are differentiable on an open interval I except possibly at a point c contained in I, if

, and

exists, and

for all x in I with x ≠ c,

then

[PLAIN]https://upload.wikimedia.org/math/8/9/9/8991dfbd9db5990224ae803c727464a7.png.

ii) $$\lim _{x \rightarrow a} (f(x) \cdot g(x)) = \lim _{x \rightarrow a} f(x) \cdot \lim _{x \rightarrow a} g(x) $$

The Attempt at a Solution

$$\begin{align}
\lim _{x \rightarrow 1} (\frac{23}{1-x^{23}}-\frac{11}{1-x^{11}})&= \lim _{x \rightarrow 1} \frac{1}{1-x}(\frac{23(1-x)}{1-x^{23}}-\frac{11(1-x)}{1-x^{11}})\\
&= \lim _{x \rightarrow 1} \frac{1}{1-x} \cdot (\lim _{x \rightarrow 1} \frac{23(1-x)}{1-x^{23}}-\lim _{x \rightarrow 1} \frac{11(1-x)}{1-x^{11}})\\
&= \lim _{x \rightarrow 1} \frac{1}{1-x} \cdot (\lim _{x \rightarrow 1} \frac{1}{x^{22}}-\lim _{x \rightarrow 1} \frac{1}{x^{10}}) \text {[Using L Hopital's Rule]} \\
&= \lim _{x \rightarrow 1} \frac{1-x^{12}}{1-x} \lim _{x \rightarrow 1} x^{22} \\
&= \lim _{x \rightarrow 1} 12x^{11} \lim _{x \rightarrow 1} x^{22} \text {[Using L Hopital's Rule]}\\
&= 12
\end{align}$$
But the correct answer is 6

 

[BvU]

Apparently somehow you are subtracting two infinities. I am used to letting Taylor series do the work for limits, but that doesn't fly here: $$\lim_{\epsilon\downarrow 0} {23\over 1-(1+\epsilon)^{23} } = \lim_{\epsilon\downarrow 0} {23\over 23\epsilon}$$ and there you go. Same with the other one. Conclusion: one more Taylor term needed and then you'll get the ${1\over2}$ to produce the book result.

 

[Ray Vickson]

Homework Statement

$$\lim _{x \rightarrow 1} (\frac{23}{1-x^{23}}-\frac{11}{1-x^{11}})$$

Homework Equations

i) For functions f and g which are differentiable on an open interval I except possibly at a point c contained in I, if

, and

exists, and

for all x in I with x ≠ c,

then

[PLAIN]https://upload.wikimedia.org/math/8/9/9/8991dfbd9db5990224ae803c727464a7.png.

ii) $$\lim _{x \rightarrow a} (f(x) \cdot g(x)) = \lim _{x \rightarrow a} f(x) \cdot \lim _{x \rightarrow a} g(x) $$

The Attempt at a Solution

$$\begin{align}
\lim _{x \rightarrow 1} (\frac{23}{1-x^{23}}-\frac{11}{1-x^{11}})&= \lim _{x \rightarrow 1} \frac{1}{1-x}(\frac{23(1-x)}{1-x^{23}}-\frac{11(1-x)}{1-x^{11}})\\
&= \lim _{x \rightarrow 1} \frac{1}{1-x} \cdot (\lim _{x \rightarrow 1} \frac{23(1-x)}{1-x^{23}}-\lim _{x \rightarrow 1} \frac{11(1-x)}{1-x^{11}})\\
&= \lim _{x \rightarrow 1} \frac{1}{1-x} \cdot (\lim _{x \rightarrow 1} \frac{1}{x^{22}}-\lim _{x \rightarrow 1} \frac{1}{x^{10}}) \text {[Using L Hopital's Rule]} \\
&= \lim _{x \rightarrow 1} \frac{1-x^{12}}{1-x} \lim _{x \rightarrow 1} x^{22} \\
&= \lim _{x \rightarrow 1} 12x^{11} \lim _{x \rightarrow 1} x^{22} \text {[Using L Hopital's Rule]}\\
&= 12
\end{align}$$
But the correct answer is 6

You cannot write
$$\lim_{x \to 1} \frac{1}{1-x} \cdot \left( \frac{23(1-x)}{1-x^{23}} - \frac{11(1-x)}{1-x^{11}} \right)$$
as
$$\lim_{x \to 1} \frac{1}{1-x} \cdot \left( \lim_{x \to 1} \frac{23(1-x)}{1-x^{23}} - \lim_{x \to 1} \frac{11(1-x)}{1-x^{11}} \right)$$
because the first factor $= \pm \infty$. Instead, take $x = 1 + h$ and expand out $x^n = (1+h)^n$ in powers of $h$.