Evaluating the given limit without using L'H rule.

Question

Evaluate $\;\lim\limits_{x\to 1} \cfrac{x^{P+1} - x(P+1) + P}{(x-1)^2} $.

The reason why I've mentioned for help in evaluating the given limit without using L'H rule is that I've already solved it using L'H rule twice (shown below) but I'm actually looking for an alter method for this question, that is without using L'H rule. Just as a curiosity to solve it in multiple ways.

Using L'H rule, this is what I've done:

$$\begin{align} & \lim_{x\to 1} \cfrac{(P+1)x^P - (P+1)}{2(x-1)} \tag{1}\\ &\lim_{x\to 1} \cfrac{P(P+1)x^{P-1}}{2} \tag{2} \\ \implies &\lim_{x\to 1} \cfrac{x^{P+1} - x(P+1) + P}{(x-1)^2}= \cfrac{P(P+1)}{2}\end{align}$$

I've used L'H rule twice here, in steps (1) and (2).

The above work was just to show what I've tried yet. I'm not sure how to move forward without using L'H rule. Any help will be greatly appreciated.

Answer 1

Set $y=x-1$, so we're dealing with $$ \lim_{y\to 0} \frac{(1+y)^{P+1}-(1+y)(P+1)+P}{y^2} = \lim_{y\to 0} \frac{(1+y)^{P+1}-1-(P+1)y}{y^2} $$ Now, the binomial theorem says $$ (1+y)^{P+1} = 1 + (P+1)y + \frac{P(P+1)}{2}y^2 + O(y^3), $$ and so on.

Answer 2

Using Taylor series built at $x=1$, you find

$$x^{P+1} - x(P+1) + P=\frac{1}{2} P (P+1) (x-1)^2+\frac{1}{6} (P-1) P (P+1) (x-1)^3+O\left((x-1)^4\right)$$ which shows, for the limit you consider not only the value but also how it is approached.

Answer 3

Write $X^{p+1}-(p+1)X+p=X(X^p-1)-p(X-1)$ factor $(X-1)$ so the numerator looks like $(X-1)[X^p+X^{p-1}+\cdots+X-p]$. Now look closely at the second factor and rewrite it

$$(X^p-1)+(X^{p-1}-1)+\cdots+(X-1)=(X-1)[(X^{p-1}+\cdots+1)+(X^{p-2}+\cdots+1)+\cdots+1]$$

So our limit is

$$\begin{align}\lim_{X\to 1}(X^{p-1}+\cdots+1)+(X^{p-2}+\cdots+1)+\cdots+1&=p+(p-1)+\cdots+1\\&=\frac{p(p+1)}{2}\end{align}$$

Answer 4

Let $f(x) = x^{p+1} -(p+1)x.$ We are looking at

$$(1)\,\,\,\,\lim_{x\to 1}\frac{f(x)-f(1)}{x-1}.$$

Now $f(x)-f(1) = \int_1^x (p+1)(t^p-1)\,dt.$ From the definition of the derivative of $t^p$ at $1,$ we have $t^p - 1 = p(t-1) +o((t-1)).$ Thus

$$f(x)-f(1) = \int_1^x (p+1)p(t-1)\,dt + \int_1^xo( (t-1))\,dt$$ $$ = (p+1)p(x-1)^2/2 +o((x-1)^2).$$

This shows $$(1) = (p+1)p/2.$$