The limit of x^1/x as x approaches infinity is equal to 1. This limit is denoted by lim_{x→∞} x^{1/x}, so the formula for the limit of x^{1/x} when x tends to infinity is given by

lim_{x→∞} x^{1/x} = 1.

Let us now find the limit of x to the power 1/x when x tends to ∞.

## Proof of Limit x^{1/x} when x→∞

Answer: The limit of x^{1/x} when x→∞ is equal to 1. |

**Explanation:**

**Step 1:**

Use the exponential formula: $x^y=e^{\ln(x^y)}$. Thus, the given limit will be

lim_{x→∞} x^{1/x} = lim_{x→∞} $e^{\ln(x^{1/x})}$

⇒ lim_{x→∞} x^{1/x} = lim_{x→∞} $e^{\frac{1}{x} \ln(x)}$ using the formula ln(x^{y}) = y ln(x).

⇒ lim_{x→∞} x^{1/x} = $e^{\lim\limits_{x \to \infty} \frac{\ln x}{x}}$ **…(∗)**

**Step 2:**

Now we compute the limit lim_{x→∞} $\dfrac{\ln x}{x}$.

Note that this has the indeterminate form ∞/∞. So using the l’Hospital’s Rule, we obtain that

lim_{x→∞} $\dfrac{\ln x}{x}$ = lim_{x→∞} $\dfrac{1/x}{1}$ = lim_{x→∞} $\dfrac{1}{x}$ = 0.

**Step 3:**

Using lim_{x→∞} ln(x)/x = 0, we deduce from** (∗)** that

lim_{x→∞} x^{1/x} = e^{0} = 1.

So the limit of x^{1/x} is equal to 1 when x approaches 0.

More Limits:

lim_{x→0} (cosx -1)/x | lim_{x→0} cosx/x

lim_{x→0} sinx/x | lim_{x→0} tanx/x

Limit of (1+ 1/x)^{x} when x→∞

## FAQs

### Q1: What is the limit of x^{1/x} when x tends to infinity?

Answer: The limit of x^{1/x} when x tends to infinity is equal to 1, that is, lim_{x→∞} x^{1/x} = 1.