Question

Evaluate:
limit x tends to 0 ( sin x/ x)^(1/x)​

Answer 1

Answer:

easy way:

limit x tends to 0 (sinx/x)

= 1

limit x tends to 0 (1/x)

= ∞

so  limit x tends to 0 ( sin x/ x)^(1/x)​

= 1^∞

=1

Step-by-step explanation:

limit x tends to 0 ( sin x/ x)^(1/x)​

= limit x tends to 0 ( (sin x)^(1/x) ) / ( x^(1/x) )​

= 0/0

use L'Hôpital's rule

= limit x tends to 0 ( ln(sin x)^(1/x) ) / ln( x^(1/x) )

= limit x tends to 0 ((1/x)(ln sin x) / (1/x)(lnx))​​​

= limit x tends to 0 ( d((1/x)(ln sinx)/dx / d((1/x)lnx/dx )

=  limit x tends to 0 ( (1/x)(1/sinx)(cosx) + (ln sinx)(-1/x^2)  / (1/x)(1/x)+(lnx)(-1/x^2))

=  limit x tends to 0 ((tanx)/x - ((ln sinx)/x^2)) / (1/x^2) - (lnx)(/x^2))

= limit x tends to 0 (x tanx - ln sinx) / (1 - lnx)

= ∞/∞

use L'Hôpital's rule

limit x tends to 0 (x tanx - ln sinx) / (1 - lnx)

= limit x tends to 0 (x sec^2 x+ tanx - (1/sinx)(cosx)) / 1/x)

= limit x tends to 0 (x sec^2 x+ tanx - (1/sinx)(cosx)) / 1/x)

= limit x tends to 0 x(x sec^2 x+ tanx - cotx)

= limit x tends to 0 x^2 sec^2 x+ xtanx - xcotx)

= limit x tends to 0 (x^2 sec^2 x+ xtanx) - limit x tends to 0 (xcotx)

= 0 - limit x tends to 0 (xcotx)

= - limit x tends to 0 (x/tanx)

=0/0

use L'Hôpital's rule

= - limit x tends to 0 (x/tanx)

= - limit x tends to 0 (-1/sec^2 x)

= 1