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+12 votes

$\lim_{x\rightarrow \infty } x^{ \tfrac{1}{x}}$ is

  1. $\infty $
  2. 0
  3. 1
  4. Not defined
asked in Calculus by Boss (41.2k points)
retagged by | 2k views

4 Answers

+24 votes
Best answer

Apply an exponential of a logarithm to the expression.

$\lim_{x \to \infty} x^{\frac{1}{x}}=\lim_{x \to \infty}\exp\left ( \log\left (x^{\frac{1}{x}} \right ) \right )=\lim_{x \to \infty}\exp\left ( \frac{\log\left ( x \right )}{x} \right )$

$\lim_{x \to \infty}\exp\left ( \frac{\log\left ( x \right )}{x} \right )$

Since the exponential function is continuous, we may factor it out of the limit.

$\lim_{x \to \infty}\exp\left ( \frac{\log\left ( x \right )}{x} \right )=\exp\left (\lim_{x \to \infty}\frac{\log\left ( x \right )}{x} \right )$

$\exp\left ( \lim_{x \to \infty}\frac{\log\left ( x \right )}{x} \right )$

Logarithmic functions grow asymptotically slower than polynomials.

Since $\log\left ( x \right )$ grows asymptotically slower than the polynomial $x$ as $x$ approaches $\infty$,

$\lim_{x \to \infty}\frac{\log\left ( x \right )}{x}=0$:

$e^{0} = 1$

Correct Answer: $C$

answered by Active (1.6k points)
edited by

here exp( Expression) means

exp(expression ) right ?

+16 votes

taking log both the sides lny=lim x->infinity 1/x*lnx

                                            =lim x->infinity lnx/x  (numerator gives finite value and denominator gives infinity and finite/infinity=0   )           



so 1 is the answer


answered by Active (5k points)
How does numerator give finite value ?

The limit of the logarithm of x when x approaches infinity is infinity
+1 vote
I have tried like this -

y=lim x->inf (x^(1/x))

Taking log on both sides we get

ln y= lim x->inf (lnx^(1/x))

ln y= lim x->inf   (lnx/x)

We do the L-H rule and differentiate

ln y =lim x->inf ((1/x))

ln y =0


answered by Active (4k points)
0 votes


We can clearly see that the graph of $y=x^{\frac{1}{x}}$ is a straight line parallel to x- axis giving a constant value of y =1 for all the values of x.

So when x$\rightarrow \infty$ then also we will get y=1.

$\therefore y= _{ x\rightarrow \infty }^{lim}x^{\frac{1}{x}} = 1$

answered by Loyal (8.3k points)

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