**Q: Find the limit as x approaches 0 from the right of (sinx)^(2/lnx).**

A: Since:

limit as x→0^{+} sinx = 0

and

limit as x→0^{+} 2/lnx = 0

We have 0^{0} which is one of our indeterminate forms.

Therefore, we let:

y = (sinx)^(2/lnx)

Take the natural log of both sides to get:

lny = ln((sinx)^(2/lnx))

Use our logarithm rules to get:

lny = (2/lnx)*ln(sinx)

lny = 2 ln(sinx) / (lnx)

And, lim x→0^{+} (2 ln(sinx) / (lnx)) = ∞/∞

so… we may apply L’Hopital’s rule!

Recall: L’Hopital’s Rule tells us that lim x→a f(x)/g(x) = lim x→a f'(x)/g'(x) [if the limit exists or is infinity]

So,

lim x→0^{+} (2 ln(sinx) / (lnx)) = lim x→0^{+} (2 * (1/sinx) * (cosx)) / (1/x))

= lim x→0^{+} (2/tanx) / (1/x)

= lim x→0^{+} (2x/tanx)

= 2 * lim x→0^{+} (x/tanx)

= 2 * 1 = 2

Therefore, we have:

lny = ln((sinx)^(2/lnx))

And, lim x→0^{+} lny = 2

So, lim x→0^{+} y = e².

Final answer: e².