Solving limits with indeterminate form

Question

I received the following problem:$$\lim_{x\to\infty}\sqrt[4]{x^4+x^3}-x$$

I cannot figure out how to tackle it. Simple substitution seems to get $\infty-\infty$. This is where I got when trying to play with it: $$=\lim_{x\to\infty}\sqrt[4]{x^4+x^3}-\frac{1}{\frac{1}{x}}$$$$=\lim_{x\to\infty}\frac{\frac{1}{x}\sqrt[4]{x^4+x^3}-1}{\frac{1}{x}}$$$$=\lim_{x\to\infty}\frac{\frac{1}{x}x\sqrt[4]{1+\frac{1}{x}}-1}{\frac{1}{x}}$$$$=\lim_{x\to\infty}\frac{\sqrt[4]{1+\frac{1}{x}}-1}{\frac{1}{x}}$$

I did see the answer - its $\frac14$. But, how do you proceed from here?

Answer 1

Continue with what you have,

$$\lim_{x\to\infty}\frac{\sqrt[4]{1+\frac{1}{x}}-1}{\frac{1}{x}} =\lim_{x\to\infty}\frac{\sqrt{1+\frac{1}{x}}-1}{\frac{1}{x}(\sqrt[4]{1+\frac{1}{x}}+1)} =\lim_{x\to\infty}\frac{\frac{1}{x}}{\frac{1}{x}(\sqrt[4]{1+\frac{1}{x}}+1)(\sqrt{1+\frac{1}{x}}+1)}=\frac14$$

Answer 2

Hint: Use the fact that $x=\sqrt[4]{x^4}$ together with the fact that $a-b=\dfrac{a^4-b^4}{a^3+a^2b+ab^2+b^3}$.

On the other hand$$\lim_{x\to\infty}\frac{\sqrt[4]{1+\frac1x}-1}{\frac1x}=\lim_{x\to0^+}\frac{\sqrt[4]{1+x}-1}x=\frac14,$$since this last limit is the derivative at $0$ of $\sqrt[4]{1+x}$.

Answer 3

Using binomial expansion, it's $\lim\limits_{x\to\infty}\left[x\left(1+\dfrac1x\right)^{1/4}-x\right]$

$=\lim_\limits{x\to\infty}\left[x\left(1+\dfrac1{4x}+o\left(\dfrac1{x^2}\right)\right)-x\right]=\lim_{x\to\infty}\left[\dfrac14+o\left(\dfrac 1x\right)\right]=\dfrac14$

Answer 4

First multiply with $\sqrt[4]{x^4+x^3}+x$ to get a difference of squares and then repeat it: \begin{align} \sqrt[4]{x^4+x^3}-x&= \frac{\left(\sqrt[4]{x^4+x^3}-x\right)\left(\sqrt[4]{x^4+x^3}+x\right)}{\sqrt[4]{x^4+x^3}+x}\\ &= \frac{\sqrt{x^4+x^3}-x^2}{\sqrt[4]{x^4+x^3}+x}\\ &= \frac{\left(\sqrt{x^4+x^3}-x^2\right)\left(\sqrt{x^4+x^3}+x^2\right)}{\left(\sqrt[4]{x^4+x^3}+x\right)\left(\sqrt{x^4+x^3}+x^2\right)}\\ &= \frac{x^4+x^3-x^4}{\left(\sqrt[4]{x^4+x^3}+x\right)\left(\sqrt{x^4+x^3}+x^2\right)}\\ &= \frac{x^3}{\left(\sqrt[4]{x^4+x^3}+x\right)\left(\sqrt{x^4+x^3}+x^2\right)}\\ &= \frac{1}{\left(\sqrt[4]{1+\frac1x}+1\right)\left(\sqrt{1+\frac1x}+1\right)}\\ &\xrightarrow{x\to\infty} \frac14 \end{align}

Answer 5

Set $1/x=h$ to find $$\lim_{h\to0^+}\dfrac{(1+h)^{1/4}-1}h$$

Now set $(1+h)^{1/4}-1=r\implies h=(1+r)^4-1=4r+O(r^2)$

Can you take it from here?