Question

Evaluate:  \(\rm \lim_{x\rightarrow ∞}{x^{2/x}} = ?\) 
1. 2
2. ∞
3. 1
4. 0

Answer 1

Correct Answer - Option 3 : 1

Concept:

L-Hospital Rule: Let f(x) and g(x) be two functions

Suppose that we have one of the following cases,

I.  \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = \frac{0}{0}\)

II. \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = \frac{\infty }{\infty }\)

Then we can apply L-Hospital Rule as:

\(\mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}\left( {\bf{x}} \right)}}{{{\bf{g}}\left( {\bf{x}} \right)}} = \mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}'\left( {\bf{x}} \right)}}{{{\bf{g}}'\left( {\bf{x}} \right)}}\)

Calculation:

Let y = \(\rm \lim_{x\rightarrow ∞}{x^{2/x}}\) 

By taking log on both sides we get.

⇒ logy = \(\rm \lim_{x\rightarrow ∞}{log(x^{2/x})}\)

⇒ \(\rm logy =\lim_{x\rightarrow ∞}\frac{2logx}{x} = \frac{\infty}{\infty}\)

So, by using L'Hospital's rule i.e. Differentiating numerator and denominator w.r.t. x. 

⇒ \(\rm logy =\lim_{x\rightarrow ∞}\frac{2/x}{1} = 0\)  

⇒ log y = 0

⇒ y = e⁰ = 1.

Hence, option 3 is the correct answer.