Posted on

Limit: Indeterminate Power

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.

Leave a Reply