AP Calculus AB and BC

Limit of x^(sin x) at 0 from the Right Is 1

The limit of x raised to the sine of x, as x approaches 0 from the right, is 1. The form is 0 to the power 0. Taking logarithms turns it into sine x times log x, which rewrites as log x over cosecant x, and L'Hopital sends that to 0. Undoing the log gives 1.

lim⁡x→0+xsin⁡x=1\lim_{x \to 0^+} x^{\sin x} = 1

Settled by taking logarithms, then rewriting as a quotient for L Hopital.

Take logarithms to reach a quotient

Both the base and the exponent are collapsing to 00, which is the indeterminate form 000^{0}. A logarithm brings the exponent down where the standard machinery can reach it.

y=xsin⁡x⇒ln⁡y=sin⁡xln⁡xy = x^{\sin x} \quad\Rightarrow\quad \ln y = \sin x \ln x

That product is 00 times −∞-\infty, still indeterminate. Move the sine into the denominator as its reciprocal so L'Hopital has a quotient to work on.

ln⁡y=ln⁡xcsc⁡x\ln y = \frac{\ln x}{\csc x}

Differentiate, then undo the logarithm

As x→0+x \to 0^{+} the top runs to −∞-\infty and the bottom to +∞+\infty, so the rule applies. Use ddxcsc⁡x=−csc⁡xcot⁡x\frac{d}{dx}\csc x = -\csc x\cot x.

lim⁡x→0+ln⁡xcsc⁡x=lim⁡x→0+1x−csc⁡xcot⁡x=lim⁡x→0+(−sin⁡xx⋅tan⁡x)=−1⋅0=0\lim_{x \to 0^{+}}\frac{\ln x}{\csc x} = \lim_{x \to 0^{+}}\frac{\frac{1}{x}}{-\csc x\cot x} = \lim_{x \to 0^{+}}\left(-\frac{\sin x}{x}\cdot\tan x\right) = -1 \cdot 0 = 0

So ln⁡y→0\ln y \to 0. The logarithm was a detour, and exponentiating brings the answer back.

lim⁡x→0+xsin⁡x=e0=1\lim_{x \to 0^{+}} x^{\sin x} = e^{0} = 1

Why only the right-hand side counts

For −π2<x<0-\frac{\pi}{2} < x < 0 the exponent sin⁡x\sin x sits strictly between −1-1 and 00, so xsin⁡xx^{\sin x} asks for a negative base raised to a non-integer power and has no real value. There is no left-hand branch approaching the origin at all.

So there is nothing to approach from that side, and the answer must be written as a one-sided limit. Presenting it as a two-sided limit claims behaviour that the function does not have.

The mistakes students make

Two arithmetic traps and one that costs marks on a free response even when the number is right.

  • Reading 000^{0} as 00 and answering 00. The form is indeterminate, and here the exponent reaches 00 fast enough that the answer is 11.
  • Answering 00 at the end. That is the limit of ln⁡y\ln y, not of yy. The final step is to exponentiate.
  • Writing the result as a two-sided limit. xsin⁡xx^{\sin x} is undefined for negative xx, so only x→0+x \to 0^{+} makes sense.

Not sure which technique a limit wants?

The Limit Method Chooser walks the decision from direct substitution through factoring, the conjugate, and L'Hopital, and says why each one applies or fails.

Frequently asked questions

What is the limit of x^(sin x) as x approaches 0 from the right?

It is 11.

Why is 0^0 indeterminate?

Because a base heading to 00 pushes the value down while an exponent heading to 00 pushes it up toward 11. Which side wins depends on the rates, so the form alone gives no answer.

Is there a two-sided limit for x^(sin x) at 0?

No. Just to the left of 00 the power is not a real number, so the function has no left-hand branch and only the right-hand limit exists.