GATE CSE 2015 Set 1 | Question: 4

Question

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

• A. $\infty $
• B. $0$
• C. $1$
• D. Not defined

Answer 1

Apply an exponential of a logarithm to the expression.

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

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

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

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Logarithmic functions grow asymptotically slower than polynomials.

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

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

$e^{0} = 1$

Correct Answer: $C$

Comments

here exp( Expression) means

exp^{(expression )} right ?

Answer 2

Caption

Comments

@SALIL KUMAR this is the best answer.

Answer 3

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

so 1 is the answer

                                                

Comments

How does numerator give finite value ?

The limit of the logarithm of x when x approaches infinity is infinity

Answer 4

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