Question

limxtends to 1 sin^-1lnx​

Answer 1

f $x = lim$$\frac{ \sin(e {}^{x - 2} - 1) }{ log(x - 1) }$

This is $\frac{0}{0}$ form so by using L Hospital rule , differentiating the numerator and denominator

⇒f (x)$= lim \frac{ \cos(e {}^{x-2} - 1)(e {}{ x - 2} )}{ (\frac{1}{x - 1}) }$

⇒f (x) $= \cos(e {}^{2 - 2}- 1)(e {}^{2 - 2})(2 - 1)$

⇒f (x) $= 1$

Answer 2

FX = sin(ex-²-1)

log (x-1)

This is 0/0 from so by using( L) hospital rule. )

differentiating the numerator and denominator.)

=> f X =lim. cos(ex-²-1) (ex-2)

1/x-2

=} f X = cost. (e-²-² -1 ) (e²-²) (2-1)

=} f X = 1

hope it is help you ✍️✍️