Find $\lim\limits_{x \to 0} \frac{ \sqrt{1+x} - 1} { \sqrt[3]{1+x} - 1}$.

Question

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.

Answer 1

$\lim_\limits{x\to 0} \frac {\sqrt {x+1} - 1}{\sqrt [3] {x+1} -1 }$

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.

$\lim_\limits{x\to 0} \frac { \frac 12 -\frac 18 x + \cdots}{\frac 13 - \frac 19 x + \cdots} = \frac 32$

Answer 2

It's easier to see if you first make the substitution $1+x = t^3$.

Then, the limit as $x \rightarrow 0$ is transformed to $$\lim_{t\rightarrow 1}\frac{(\sqrt{t})^3-1}{(\sqrt{t})^2-1}.$$

Answer 3

Since there's a square root in the numerator and a cube root in the denominator, and $\operatorname{lcm}(2, 3) = 6$, make the substitution $t = \sqrt[6]{1+x}$, so the rational expression looks like (try this yourself before revealing):

$$ \frac{\sqrt{1+x} - 1}{\sqrt[3]{1+x} - 1} = \frac{t^3 - 1}{t^2 - 1} $$

Each of the expressions in the numerator and denominator can be factored using instances of the identity $$ a^n - b^n = (a - b)\,(a^{n-1} + a^{n-2}b + \dots + ab^{n-2} + b^{n-1}) $$ with $a = t$, $b = 1$, and either $n=3$ or $n=2$.

In other words,

$$ \frac{t^3 - 1}{t^2 - 1} = \frac{(t - 1)(t^2 + t + 1)}{(t - 1)(t + 1)} = \frac{t^2 + t + 1}{t + 1}, $$

where the final equality holds for all $t \neq 1$.

Now, $t \to 1$ as $x \to 0$, so we need only understand values near $t = 1$ but not at that point, so the cancellation in the previous calculation is valid. Thus, the limit can be evaluated by simply plugging in $t = 1$:

$$ \lim_{x \to 0} \frac{\sqrt{1+x} - 1}{\sqrt[3]{1+x} - 1} = \lim_{t \to 1} \frac{t^2 + t + 1}{t + 1} = \frac{1 + 1 + 1}{1 + 1} = \frac32. $$

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By the way, all of this can be done without explicitly mentioning the new variable $t$ (just reproduce each expression with the corresponding radicals involving $1+x$), but it's way more messy.

Answer 4

Along the lines of the other answers, I propose to use the unifying substitution $$1+x = y^6.$$ The reason for the sixth power is because it is the least common multiple of $2$ and $3$, thus it is the lowest power that will remove all square and cube roots. Hence $$\frac{\sqrt{1+x} - 1}{\sqrt[3]{1+x} - 1} = \frac{y^3 - 1}{y^2 - 1} = \frac{(y-1)(y^2+y+1)}{(y-1)(y+1)} = \frac{y^2 + y + 1}{y + 1},$$ and as $x \to 0$, we have $y \to 1$, which immediately yields the desired answer of $3/2$.

Answer 5

Using that

$$A^2-1=(A-1)(A+1)$$

$$A^3-1=(A-1)(A^2+A+1)$$

we have

$$\sqrt{1+x} - 1=\frac{x}{\sqrt{1+x} + 1}$$

$$\sqrt[3]{1+x} - 1=\frac{x}{\sqrt[3]{(1+x)^2} +\sqrt[3]{1+x}+1}$$

and therefore

$$ \require{cancel} \frac{ \sqrt{1+x} - 1} { \sqrt[3]{1+x} - 1}=\frac{\cancel x}{\sqrt{1+x} + 1}\frac{\sqrt[3]{(1+x)^2} +\sqrt[3]{1+x}+1}{\cancel x}=$$

$$=\frac{\sqrt[3]{(1+x)^2} +\sqrt[3]{1+x}+1}{\sqrt{1+x} + 1}\to \frac 32$$

Answer 6

Your question is Evaluate : $$\lim_{x\to 0}\frac{\sqrt{1+x}-1}{\sqrt[3]{1+x}-1}$$.

Ans) First of all this above limit is of the form of $\frac{0}{0}$ when $x\to 0$. Now to solve this limit I will apply here L-Hopital's rule.

Let $$L=\lim_{x\to 0} \frac{\sqrt{1+x}-1}{\sqrt[3]{1+x}-1}$$, where $f(x)=\sqrt{1+x}-1$ and $g(x)=\sqrt[3]{1+x}-1$

$$\implies L=\lim_{x\to 0}\frac{f'(x)}{g'(x)} \implies L= \lim_{x\to 0}\frac{\frac{1}{2\sqrt{1+x}}}{\frac{1}{3(1+x)^{\frac{2}{3}}}}$$[after applying L-Hopital's rule].

Finally $L= \frac{3}{2}$.