Limit of $n$th root of exponentials

Question

I'm struggling with this problem:

$$\lim_{x\to \infty} \sqrt[x]{(7^x - 6^x + x^3)} = \text{?}$$

Although I've tried to apply the squeeze theorem, as:

$$\sqrt[x]{7^x - 6^x} \leq \sqrt[x]{7^x - 6^x + x^3} \leq \sqrt{7^x-6^x}(\sqrt[x]{x})^3 $$

I'm not sure if this is correct.

Answer 1

$$\lim_{x\to \infty} \sqrt[x]{(7^x - 6^x + x^3)} = \lim_{x\to \infty} 7 \underbrace{\sqrt[x]{ 1-\left( \frac6 7 \right)^x + \frac{x^3}{7^x} }}_{ \to 1} = 7$$

Answer 2

Simply note that

$$\sqrt[x]{7^x - 6^x + x^3}=7\cdot\sqrt[x]{1 - \left(\frac67\right)^x + \frac{x^3}{7^x}}\to 7\cdot \left(1-0+0\right)^0=7$$

Answer 3

$$\lim_{x\rightarrow+\infty}\left(7^x-6^x+x^3\right)^{\frac{1}{x}}=7\lim_{x\rightarrow+\infty}\left(1-\left(\frac{6}{7}\right)^x+\frac{x^3}{7^x}\right)^{\frac{1}{x}}=7$$