Question

limx->0+ xsinx/ln(1+x^2)

Answer 1

In the form \[\frac{ 0 }{ 0 }\] try I'Hop

Answer 2

have you tried that @obey_briks ?

Answer 3

no! i have no idea what that is

Answer 4

because the limits yields the form 0/0

use l'hopital's rule; differential the numerator and the denominator , then take the limit again

Answer 5

note differentiate the numerator separately then differentiate the denominator separately

Answer 6

we'll do the numerator part.
\[x \sin x \rightarrow x \cos x + \sin x\]

Answer 7

and the derivative of the denominator

\[\frac{\text d}{\text dx}\ln(1+x^2)=\frac{2x}{1+x^2}\]

Answer 8

without L'Hopital try to make use of these properties
lim x->0 sin(x)/x = 1
lim x^2->0 log(1+x^2)/x^2 = 1
lim a->0 f(a)*g(a) = lim a->0 f(a) * lim a->0 g(a)

Answer 9

do you know this? \[\lim_{x \rightarrow 0}\frac{ e^x - 1 }{ x }=1\]

Answer 10

that is same as
lim x->0 log(1+x)/x = 1

Answer 11

@obey_briks how about listing a few forms on limits, like the one posted above, so we may have something so start with.

Answer 12

or,, simply,,
you can convert it into the form of 1^infinity by expressing as

1/limx->0 (ln (1+x^2)^(1/xsinx))

=>ans would directly be

x->0 e^-(x^2 /xsin x) => e^(-x/sinx) = ....

Answer 13

mine is just an alternative manipulation of what @experimentX said,,both are same..