# Limits - Calculating $\lim\limits_{x\to 1} \frac{x^a -1}{x-1}$, where $a \gt 0$, without using L'Hospital's rule

Calculate $$\displaystyle\lim\limits_{x\to 1} \frac{x^a -1}{x-1}$$, where $$a \gt 0$$, without using L'Hospital's rule.

I'm messing around with this limit. I've tried using substitution for $$x^a -1$$, but it didn't work out for me.

I also know that $$(x-1)$$ is a factor of $$x^a -1$$, but I don't know where to go from here.

EDIT: Solved it, posting here for future generations :)

a) We can write $$x^a$$ as $$e^{a\ln x}$$ ending up with $$\lim\limits_{x\to 1} \frac{e^{a\ln x} -1}{x-1}$$

b) Multiplying by $$\frac{a\ln x}{a\ln x}$$ we end up with: $$\lim\limits_{x\to 1} \frac{e^{a\ln x} -1}{a\ln x} \cdot \frac{\ln x}{x-1} \cdot a$$

c) Now we just have to show that the first 2 limits are equal 1, and $$\lim\limits_{x\to 1} a = a$$

• Welcom to MSE. What are you assuming about $a$? – José Carlos Santos Nov 15 '18 at 10:54
• I don't think there is much you can do without an actual definition of $x^a$. Once you have decided on one, however, you should have more to go on. – Arthur Nov 15 '18 at 10:56
• I only know that 'a' is greater than 0 real number. I know how to solve for natural numbers, but I'm stuck for when it can be any positive real number. – Bartosz Nov 15 '18 at 10:58

Hint

Make life simpler using $$x=1+y$$ which makes $$\lim\limits_{x\to 1} \frac{x^a -1}{x-1}=\lim\limits_{y\to 0} \frac{(1+y)^a -1}{y}$$

Now, use the binomial theorem or Taylor expansion.

• Hi Claude. The generalized binomial theorem is fine. Taylor is nearly tantamount to L'Hospital. – Yves Daoust Nov 15 '18 at 11:14

Do the substitution $$y=a\log x$$ (natural logarithm). Then $$x=e^{y/a}$$, so the limit becomes $$\lim_{y\to0}\frac{e^y-1}{e^{y/a}-1}= \lim_{y\to0}a\frac{e^y-1}{y}\frac{y/a}{e^{y/a}-1}$$ Since $$\lim_{y\to0}\frac{e^y-1}{y}=1$$ we can conclude that $$\lim_{x\to1}\frac{x^a-1}{x-1}=a$$

• Thank you, I also managed to solve it by myself, but this might be more elegant solution. – Bartosz Nov 15 '18 at 13:07

This is literally the definition of the derivative of $$x^a$$ at $$x=1$$, so L'Hôpital's rule would have been circular anyway. We'll have to differentiate, viz. $$y=x^a\implies \ln y=a\ln x\implies \frac{y'}{y}=\frac{a}{x}\implies y'=ax^{a-1}\implies y'\left.\right|_{x=1}=a.$$

• The problem is I'm not allowed to use derivatives to solve my problems yet. – Bartosz Nov 15 '18 at 11:20

mIf $$f(x)=x^a$$, $$\;\lim\limits_{x\to 1}\dfrac{x^a-1^a}{x-1}\;$$ is the definition of $$f'(1)$$. So the limit is $$a\,x^{a-1}\biggm|_{x=1}=\color{red}a.$$

For $$a$$ rational, let $$a=\dfrac pq$$. We set $$x=t^q$$, and

$$\frac{x^{p/q}-1}{x-1}=\frac{t^p-1}{t^q-1}=\frac{\dfrac{t^p-1}{t-1}}{\dfrac{t^q-1}{t-1}}$$

which tends to $$\dfrac pq$$.

For irrational $$a$$, the result could be extended using continuity, but this is more technical and depends on your definition of the powers.

You can use Bernoulli's ineq (which can be proved by am-gm and by continuity argument, so minimal calculus knowledge i think) and the squeeze theorem. You can assume wlog that $$a\geq 1$$, or otherwise, if $$0, take $$X=x^a$$, so $$\lim_{x\to 1}\frac{x^a-1}{x-1}=\lim_{X\to 1}\frac{X-1}{X^{\frac1a}-1}=\frac{1}{\lim_{X\to 1}\frac{X^{\frac1a}-1}{X-1}}.$$ So, if $$\lim_{x\to 1}\frac{x^a-1}{x-1}=a$$ for $$a\geq 1$$ holds, we also have $$\lim_{x\to 1}\frac{x^a-1}{x-1}=a$$ for $$0 too.

Suppose that $$a\geq 1$$. Let $$x=1+y$$. Wlog, we may say that $$y>-1$$ is a small real number. For $$a\geq 1$$, Bernoulli's ineq gives $$x^a-1=(1+y)^a-1 \geq ay=a(x-1).$$ So, $$\frac{x^a-1}{x-1}\geq a$$ if $$y>0$$ (or $$x>1$$). (The ineq is reversed for $$y<0$$, or $$x<1$$.)

Now write $$x=\frac{1}{1-z}$$ for some small real number $$z<\frac1a$$ (that is, $$z=y/(1+y)$$). By Bernoulli's ineq, $$(1-z)^a\geq 1-az$$, so $$x^a-1=\frac{1}{(1-z)^a}-1\leq \frac{1}{1-az}-1=\frac{az}{1-az}.$$ That is, if $$y>0$$ (i.e. $$x>1$$), we get $$z>0$$ and $$\frac{x^a-1}{x-1}=\frac{\frac1{(1-z)^a}-1}{\frac1{1-z}-1}\leq \frac{az}{1-az}\left(\frac{1-z}{z}\right)=\frac{a(1-z)}{1-az}.$$ The ineq is reversed if $$y<0$$ (i.e., $$x<1$$ and $$z<0$$). Therefore, $$a\leq \frac{x^a-1}{x-1}\leq \frac{a(1-z)}{1-az}=\frac{a}{x}\left(\frac{1}{1-a\left(1-\frac1x\right)}\right)=\frac{a}{x-a(x-1)}$$ if $$x>1$$.

Similarly, for $$a\geq1$$ and $$0, we have $$\frac{a}{x-a(x-1)}\leq \frac{x^a-1}{x-1}\leq a.$$ Taking $$x\to 1$$ and using the squeeze thm, we have $$\lim_{x\to 1}\frac{x^a-1}{x-1}=a$$ for $$a\geq 1$$.