how to prove by recurrency a limit

Question

I have no idea how to go about this.

prove by recurrence that $\lim_{x->0}\frac{(ax+1)^{n}-1}{x}=na$

Thanks.

Answer 1

$(ax+1)^{n+1}-(ax+1)^n =(ax+1)^{n}((ax+1)-1) =ax(ax+1)^{n} $ so $\dfrac{(ax+1)^{n+1}-1}{x}-\dfrac{(ax+1)^n-1}{x} =\dfrac{(ax+1)^{n+1}-1-((ax+1)^n-1)}{x} =\dfrac{(ax+1)^{n+1}-(ax+1)^n)}{x} =a(ax+1)^n $.

Since $\lim_{x \to 0} (ax+1)^n =1$, $\lim_{x \to 0} \dfrac{(ax+1)^{n+1}-1}{x}-\lim_{x \to 0}\dfrac{(ax+1)^n-1}{x} =a $.

Starting with $n=1$, $\lim_{x \to 0}\dfrac{(ax+1)^1-1}{x} =\lim_{x \to 0}a =a$, so by induction, $\lim_{x \to 0}\dfrac{(ax+1)^n-1}{x} =na $.

Answer 2

Note that

$$\lim_{x\to0}\frac{(ax+1)^{n}-1}{x}=\lim_{x\to0}\,\frac{f(x)-f(0)}{x-0}=f'(0)=na(a0+1)^{n-1}=na$$

Answer 3

Just use a high-school identity: $$(ax+1)^n-1=(ax+\not1-\not1)\bigl((ax+1)^{n-1}+\dots+(ax+1)+1\bigr), $$ $$\text{so }\qquad\frac {(ax+1)^n-1}{x} = a\bigl((\underbrace{ax+1)^{n-1}+\dots+(ax+1)+1}_{n \text{ terms}}\bigr)\quad\text{tends to }an.$$