Lim x - 0- xxx-1-p and lim x - 0- xxx-1-q- then p-q is equal to

The correct option is A 2lim_x → 0^+(x)^(x^x 1)=p Let I=x^x ∴ log I=lim_x → 0^+x lnx =lim_x → 0^+lnx/1/x=0 ∴ I=1 log p = lim_x → 0^+(x^x 1)lnx =lim_x → 0^+(e ...

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Byju's Answer

Standard XII

Mathematics

Continuity of a Function

lim x → 0+ xx...

Question

$l i m_{x \to 0^{+}} x^{(x^{x} - 1)} = p a n d l i m_{x \to 0^{+}} x^{x^{x}} - 1 = q$ , then p – q is equal to

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Solution

The correct option is A 2
${lim}_{x \to 0^{+}} (x)^{(x^{x} - 1)} = p$
Let $I = x^{x}$
$∴ l o g I = {lim}_{x \to 0^{+}} x l n x = {lim}_{x \to 0^{+}} \frac{l n x}{\frac{1}{x}} = 0 ∴ I = 1 l o g p = {lim}_{x \to 0^{+}} (x^{x} - 1) l n x = {lim}_{x \to 0^{+}} (e^{x l n x} - 1) l n x {lim}_{x \to 0^{+}} \frac{(e^{x l n x} - 1)}{x l n x} (x l n x) l n x {lim}_{x \to 0^{+}} \frac{(l n x)^{2}}{\frac{1}{x}} = 0$
$∴$ p = 1
$q = l i m_{x \to 0^{+}} (x^{x^{x}} - 1) = l i m_{x \to 0^{+}} (x^{1} - 1) = - 1 ∴ q = - 1 ∴ p - q = 2$

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limx→0+x(xx−1)=p and limx→0+xxx−1=q, then p – q is equal to

The correct option is A 2limx→0+(x)(xx−1)=p Let I=xx ∴ log I=limx→0+xlnx=limx→0+lnx1x=0∴ I=1log p=limx→0+(xx−1)lnx=limx→0+(exlnx−1)lnxlimx→0+(exlnx−1)xlnx(xlnx)lnxlimx→0+(lnx)21x=0 ∴ p = 1 q=limx→0+(xxx−1)=limx→0+(x1−1)=−1∴ q=−1∴ p−q=2

$l i m_{x \to 0^{+}} x^{(x^{x} - 1)} = p a n d l i m_{x \to 0^{+}} x^{x^{x}} - 1 = q$ , then p – q is equal to