limx→∞ x(1/x) =

Question

limx→∞ x(1/x) =  (A) 1 (B) ∞ (C) 0 (D) none of these

Tardigrade Admin · Accepted Answer

Let y = limx→∞ x(1/x) .....(i) Taking log in (i) on both sides, we get lim x→ ∞ (1/x ) log x Applying L'-Hospital's Rule, we get log y = lim x→ ∞ ((1/x)/1) = lim x→ ∞ (1/x) =0 or y = e0 = 1

Q. $lim_{x \to \infty} x^{\frac{1}{x}} =$

1

$\infty$

0

$n o n e o f t h ese$

Let $y = lim_{x \to \infty} x^{\frac{1}{x}}$ .....(i)
Taking log in (i) on both sides, we get
$lim_{x \to \infty} \frac{1}{x} lo g x$
Applying L'-Hospital's Rule, we get
$lo g y = lim_{x \to \infty} \frac{\frac{1}{x}}{1} = lim_{x \to \infty} \frac{1}{x} = 0$
or $y = e^{0} = 1$

Q. limx→∞​xx1​=

1

∞

0

noneofthese

Let y=limx→∞​xx1​ .....(i) Taking log in (i) on both sides, we get limx→∞​x1​logx Applying L'-Hospital's Rule, we get logy=limx→∞​1x1​​=limx→∞​x1​=0 or y=e0=1

Q. $lim_{x \to \infty} x^{\frac{1}{x}} =$

$\infty$

$n o n e o f t h ese$

Let $y = lim_{x \to \infty} x^{\frac{1}{x}}$ .....(i)
Taking log in (i) on both sides, we get
$lim_{x \to \infty} \frac{1}{x} lo g x$
Applying L'-Hospital's Rule, we get
$lo g y = lim_{x \to \infty} \frac{\frac{1}{x}}{1} = lim_{x \to \infty} \frac{1}{x} = 0$
or $y = e^{0} = 1$