Limit theorem problem In my calculus book, there are two theorems :
(i)$$\lim_{x \rightarrow a}x^n=a^n$$ and (ii)
$$\lim_{x\rightarrow a}\frac{x^n-a^n}{x-a}=na^{n-1}$$
$\forall x\in R^+ \forall n\in R.$
In my textbook, the proof of above theorems are given $\forall n\in N$. And told that "We will accept the theorem $\forall n\in R$ and apply."
I am a beginner in calculus.So, I can't prove these theorems $\forall n\in R$, and also on other websites, I can't find the proof of above theorems. So, tell me how to prove these theorems. Also give me name of books which have proofs of above theorems and which is best one for studying Analysis.THANKS
 A: You have to understand that the first result $\lim_{x \to a}x^{n} = a^{n}$ is a consequence of the second result $\lim_{x \to a}\dfrac{x^{n} - a^{n}}{x - a} = na^{n - 1}$. This follows in the following manner: $$\lim_{x \to a}x^{n} = \lim_{x \to a}x^{n} - a^{n} + a^{n} = \lim_{x \to a}(x - a)\left(\frac{x^{n} - a^{n}}{x - a}\right) + a^{n} = 0\cdot na^{n - 1} + a^{n} = a^{n}$$ The proof of second result is bit longer. I am pasting the proof from my blog post (which has proofs for other standard limit theorems also) below:
In case $n$ is positive integer the result follows easily by Binomial theorem as follows: $$\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} = \lim_{h \to 0}\frac{(a + h)^{n} - a^{n}}{h} = \lim_{h \to 0}\frac{(a^{n} + na^{n - 1}h + \cdots) - a^{n}}{h} = na^{n - 1}$$ If $n = 0$ then clearly the numerator vanishes and limit is $0$ so the formula is true in this case also. If $n$ is negative integer $n = -m$ so that $m$ is a positive integer, then $$\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} = \lim_{x \to a}\frac{a^{m} - x^{m}}{(x - a)x^{m}a^{m}} = -\frac{ma^{m - 1}}{a^{2m}} = -ma^{-m - 1} = na^{n - 1}$$ To handle the case when $n = p/q$ is a fraction we restrict to case when $a > 0$ and $p, q$ are positive integers. First we note that $\lim_{x \to a}(x^{q} - a^{q})/(x - a) = qa^{q - 1}$ so that the ratio $(x^{q} - a^{q})/(x - a)$ is bounded but away from zero when $x$ is in certain neighbourhood of $a$ (and since $\lim_{x \to a}x^{q} = a^{q}$ it follows that $x^{q}$ is in a certain neighbourhood of $a^{q}$). Replacing $x$ by $x^{1/q}$ and $a$ by $a^{1/q}$ we note that $(x - a)/(x^{1/q} - a^{1/q})$ is bounded but away from zero if $x$ is in a certain neighborhood of $a$. Thus we see that $(x^{1/q} - a^{1/q})/(x - a)$ is also bounded and away from zero when $x \to a$. It follows that $\lim_{x \to a}x^{1/q} = a^{1/q}$. We can now see that
\begin{align}
\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} &= \lim_{x \to a}\frac{x^{p/q} - a^{p/q}}{x^{1/q} - a^{1/q}}\cdot \frac{x^{1/q} - a^{1/q}}{x - a}\notag\\
&= \lim_{y \to b}\frac{y^{p} - b^{b}}{y - b}\cdot\frac{y - b}{y^{q} - b^{q}}\text{ (putting }y = x^{1/q}, b = a^{1/q})\notag\\
&= pb^{p - 1}/qb^{q - 1} = \frac{p}{q}b^{p - q} = na^{n - 1}\notag
\end{align}
If $n$ is negative rational then the proof follows exactly along the same lines as the case for $n$ being a negative integer. In the above proof notice that it is absolutely essential that we establish that $\lim_{x \to a}x^{1/q} = a^{1/q}$ first and it requires a little bit of ingenuity in achieving this.
When $n$ is irrational then the definition of $x^{n}$ is dependent on the definitions of exponential and logarithm functions. In this case $a$ must be positive otherwise $a^{n}$ is not defined. We have $x^{n} = \exp(n\log x)$ and $a^{n} = \exp(n\log a)$. Let $\log x = y$ and $\log a = b$ so that $x \to a$ implies $y \to b$. We now have
\begin{align}
\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} &= \lim_{y \to b}\frac{e^{ny} - e^{nb}}{e^{y} - e^{b}}\notag\\
&= \lim_{y \to b}\frac{e^{ny} - e^{nb}}{n(y - b)}\cdot\frac{n(y - b)}{e^{y} - e^{b}}\notag\\
&= n\lim_{h \to 0}\frac{e^{n(b + h)} - e^{nb}}{nh}\cdot\frac{h}{e^{b + h} - e^{b}}\notag\\
&= n \lim_{h \to 0}\frac{e^{nb}(e^{nh} - 1)}{nh}\cdot\frac{h}{e^{b}(e^{h} - 1)}\notag\\
&= ne^{nb}/e^{b} = ne^{b(n - 1)} = n(e^{b})^{n - 1} = na^{n - 1}\notag
\end{align}
The above proofs based on definitions of exponential and logarithm functions are not suitable for a beginner learning limits for the first time because they depend upon the higher level concepts of integral and derivative. Also these proofs make use of properties of exponential and logarithmic in an implicit fashion (these can not be proved here because of the constraint to keep the post to a reasonable length). However when one has got an understanding of derivative and integral then it is better to revisit these proofs which justify the use of fundamental logarithmic and exponential in solving various limit problems.
