Find the limit of the following function

Question

Find $$\lim_{x \to 0}\frac{(1+x)^{1/m}-(1-x)^{1/m}}{(1+x)^{1/n}-(1-x)^{1/n}}$$

I tried solving, but dividing by $x$ doesn't help in this case. Please help me to solve this question. Thanks in advance.

Answer 1

You can use binomial expansion: For the numerator $$(1+x)^{1/m}-{(1-x)}^{1/m}=\frac{2}{m}x+\frac{2}{m}\cdot(\frac{1}{m}-1)\cdot(\frac{1}{m}-2)x^2+\cdots$$ and denominator $$(1+x)^{1/n}-{(1-x)}^{1/n}=\frac{2}{n}x+\frac{2}{n}\cdot(\frac{1}{n}-1)\cdot(\frac{1}{n}-2)x^2+\cdots$$ Now, $$\lim_{x \to 0}\frac{\frac{2}{m}x+\frac{2}{m}\cdot(\frac{1}{m}-1)\cdot(\frac{1}{m}-2)x^2+\cdots}{\frac{2}{n}x+\frac{2}{n}\cdot(\frac{1}{n}-1)\cdot(\frac{1}{n}-2)x^2+\cdots}=\frac{n}{m}$$ Now, dividing by x and evaluating at $x=0$, we get: $$\lim_{x \to 0}\frac{(1+x)^{1/m}-(1-x)^{1/m}}{(1+x)^{1/n}-(1-x)^{1/n}}=\frac{n}{m}$$

Answer 2

Using binomial series, it's

$$\lim_{x \to 0}\frac{(1+x)^{1/m}-(1-x)^{1/m}}{(1+x)^{1/n}-(1-x)^{1/n}}$$

$$=\lim_{x \to 0}\frac{(1+\frac1mx)-(1-\frac1mx)}{(1+\frac1nx)-(1-\frac1nx)}$$

$$=\lim_{x \to 0}\frac{\frac2mx}{\frac2nx}$$

$$=\frac nm$$

Answer 3

A little algebra often helps. Add subtract $1$ in both numerator and denominator and then divide each by $x$ and transform the expression under limit to $$\frac{f(x, m) + f(-x, m)} {f(x, n) + f(-x, n)} $$ where $$f(a, b) =\frac{(1+a)^{1/b}-1}{a}$$ Using the formula $$\lim_{x\to a} \frac{x^r-a^r} {x-a} =ra^{r-1}$$ it follows that $f(a, b) \to 1/b$ as $a\to 0$. Hence the desired limit is $(1/m+1/m)/(1/n+1/n)=n/m$.