I'm having trouble finding $\lim\limits_{x \to 0} \frac{ \sqrt{1+x} – 1} { \sqrt[3]{1+x} – 1}$. Here's my attempt:
$$
\lim_{x \to 0} \frac{\sqrt{1+x} – 1}{\sqrt[3]{1+x} – 1}
= \lim_{x \to 0} \frac{\sqrt{1+x} – 1}{\sqrt[3]{1+x} – 1}
\cdot \frac{\sqrt{1+x} + 1}{\sqrt{1+x} + 1}
= \lim_{x \to 0} \frac{x}{(\sqrt[3]{1+x} – 1)
\cdot (\sqrt{1+x} + 1)}
$$
I'm having trouble getting rid of the given expression in the denominator. My professor mentioned multiplying with
$$
\frac{\sqrt[3]{(1+x)^2} + \sqrt[3]{1+x} + 1}
{\sqrt[3]{(1+x)^2} + \sqrt[3]{1+x} + 1}
$$
because of the factorization of $a^n-b^n$ (???) but even when I do that I don't get the correct result, which is $\frac{3}{2}$.
Any alternative ways to solve this would be appreciated, if someone could explain the professors logic it would also be of great help. Thanks.
Best Answer
$\lim_\limits{x\to 0} \frac {\sqrt {x+1} - 1}{\sqrt [3] {x+1} -1 }$
As you said you multiply by $\frac {\sqrt {x+1} + 1}{\sqrt {x+1} + 1}$ to clear the radical in the numerator
$\lim_\limits{x\to 0} \frac {\sqrt {x+1} - 1}{\sqrt [3] {x+1} -1 }\frac {\sqrt {x+1} + 1}{\sqrt {x+1} + 1} = \frac {x}{(\sqrt [3] {x+1} -1)(\sqrt {x+1} + 1) }$
As your prof. suggested we multiply by $\frac {(\sqrt [3] {x+1})^2 + \sqrt [3] {x+1} + 1}{(\sqrt [3] {x+1})^2 + \sqrt [3] {x+1} + 1}$ as in the denominator $(\sqrt [3] {x+1} - 1)((\sqrt [3] {x+1})^2 + \sqrt [3] {x+1} + 1) = (\sqrt [3] {x+1})^3 - 1 = x$
$\lim_\limits{x\to 0} \frac {x((\sqrt [3] {x+1})^2 + \sqrt [3] {x+1} + 1)}{x(\sqrt {x+1} + 1) }$
$x$ factors in the numerator and denominator cancel and then set $x$ to 0.
$\lim_\limits{x\to 0} \frac {(\sqrt [3] {1})^2 + \sqrt [3] {1} + 1)}{\sqrt {1} + 1 } = \frac {3}{2}$
But, if you are looking for an alternative.
The binomial theorem says $(a+b)^n = a^n + na^{n-1}b + \frac {n(n-1)}{2}a^{n-2}b^2 + \cdots$ While I am sure you have seen this for integer exponents, it works for non-integers just the same.
Apply the binomial theorem to $(1+x)^\frac 12 = 1 + \frac 12 x - \frac 18 x^2 + \cdots$
and to $(1+x)^\frac 13 = 1 + \frac 13 x - \frac 19 x^2 + \cdots$
Substituting the power series into the limit and simplifying.
$\lim_\limits{x\to 0} \frac { \frac 12 -\frac 18 x + \cdots}{\frac 13 - \frac 19 x + \cdots} = \frac 32$