AP Calculus AB and BC

Limit of ln(sin x)/ln(x) at 0 from the Right

The limit of ln of sin x, over ln x, as x approaches 0 from the right is 1. Both logarithms run to negative infinity, and since sin x behaves like x near 0 their ratio tends to 1. The limit is one sided because sin x must stay positive.

lim⁡x→0+ln⁡(sin⁡x)ln⁡x=1\lim_{x \to 0^+} \frac{\ln(\sin x)}{\ln x} = 1

Settled by L'Hopital's rule, or the leading-order comparison.

Why it must be one sided

For x<0x < 0 near the origin sin⁡x\sin x is negative, so ln⁡(sin⁡x)\ln(\sin x) does not exist. The two-sided limit is not merely hard, it is meaningless, which is why the page states a side.

Both parts run to negative infinity

As x→0+x \to 0^{+} both sin⁡x\sin x and xx approach 00 from above, so both logarithms run to −∞-\infty and the form is ∞∞\frac{\infty}{\infty} in size.

lim⁡x→0+ln⁡(sin⁡x)ln⁡x  =H  lim⁡x→0+cos⁡xsin⁡x1x=lim⁡x→0+xcos⁡xsin⁡x=1\lim_{x \to 0^{+}}\frac{\ln(\sin x)}{\ln x} \;\overset{\text{H}}{=}\; \lim_{x \to 0^{+}}\frac{\frac{\cos x}{\sin x}}{\frac{1}{x}} = \lim_{x \to 0^{+}}\frac{x\cos x}{\sin x} = 1

The last step is the standard limit again, with cos⁡x→1\cos x \to 1. The underlying reason is simply that sin⁡x\sin x and xx agree to leading order, so their logarithms differ by a vanishing amount.

The mistakes students make

  • Reporting a two-sided limit. It does not exist, because the function is undefined for x<0x < 0.
  • Cancelling the logarithms to get sin⁡xx\frac{\sin x}{x}. There is no such rule, though the answer happens to coincide.

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 ln(sin x)/ln(x) as x approaches 0 from the right?

It is 11.

Why only from the right?

For x<0x < 0 near 00, sin⁡x\sin x is negative and its logarithm is undefined.