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 00 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.
