Question

Evaluate the limit as x approaches 0 of (1 - x^(sin(x)))/(x*log(x))

aka: \(\frac{1-x^{sin(x)}}{x*log(x)}\)

Please include steps/explanation.

Answer 1

@ganeshie8 @Michele_Laino

Answer 2

use l'Hospital rrule

Answer 3

\(\LARGE \frac{1-x^{sin(x)}}{x*log(x)}\)

Answer 4

\(\large \lim_{x \to 0^{+}} \frac{1- x^{x} }{ \log( x^x) } =_{LH} \lim_{x \to 0^{+}} \frac{0 -x^x( 1 + \log (x)) }{1 + \log (x) } \\ = \large \lim_{x \to 0^{+}} (-x^x) = \large - \lim_{x \to 0^{+}} (x^x) = -1\)

Answer 5

@Er.Mohd.AMIR correct?

Answer 6

@sleepyhead314 I know you know the answer ;)

Answer 7

OK I can solve the problem

Answer 8

cannot directly use l'hopital

Answer 9

yea, before that I had this;

\(\large \lim_{x \to 0^{+}} \frac{1- x^{\sin x} }{x \log x } \\~\\ \large = \lim_{x \to 0^{+}} \frac{1- x^{\sin x} }{ \log( x^x) } \\~\\ \large = \lim_{x \to 0^{+}} \frac{1- x^{x} }{ \log( x^x) } ~~ \normalsize{\text{ substituting x for sin x } } \\~\\ \large = \frac{\lim_{x \to 0^{+}} (1) - \lim_{x \to 0^{+}} \left( x^{x}\right) }{ \log( \lim_{x \to 0^{+}}x^x) } = \frac{1-1}{\log(1)} = \frac{0}{0}\)

Answer 10

I know that you know
that you have absolutely no idea what that means

Answer 11

or do I (;

Answer 12

wrong diff of numerator.

Answer 13

check my solution

Answer 14

@hartnn :)

Answer 15

calculus problem lol

Answer 16

@TheSmartOne did you tag Hartnn for help or for CoC enforcement? Cause I can help with the question.

Answer 17

for help :)

Answer 18

@ganeshie8 does my solution contains error? do let me know

Answer 19

if you can help @just_one_last_goodbye , go for it :)

Answer 20

working it out on my paper :)

Answer 21

sure, tell me what you get :)