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

1. $\infty$
2. 0
3. 1
4. Not defined
in Calculus
retagged | 2.4k views

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: $C$

by Active (1.6k points)
edited
+3

here exp( Expression) means

exp(expression ) right ?

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   )

ln(y)=0

y=e0=1

by Active (5.1k points)
+1
How does numerator give finite value ?

The limit of the logarithm of x when x approaches infinity is infinity
by (55 points)
+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

y=e^(0)

y=1
by Active (4.7k points)

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.

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

by Boss (23.7k points)