AP Calculus AB and BC

Limit of ln(1 + x^2)/x^2 at 0 Is 1

Q: Is substituting u = x^2 allowed inside a limit?

Yes, provided u \to 0 as x \to 0, which it does, and the limit in u exists. Both hold here.

Q: Does the side matter for this one?

No. Since x^{2} is never negative, the expression takes the same values at x and -x, so the two one-sided limits are identical.

The limit of the natural log of 1 plus x squared, over x squared, as x approaches 0 is 1. Setting u equal to x squared turns the quotient into log of 1 plus u over u, the standard limit with value 1. The substitution is legal because u tends to 0 as x does.

\lim_{x \to 0} \frac{\ln\left(1+x^{2}\right)}{x^{2}} = 1

Settled by substituting u = x squared into the standard logarithm limit.

One substitution does everything

The expression that appears inside the logarithm and the expression in the denominator are the same, $x^{2}$ . Call it $u$ .

\lim_{x \to 0}\frac{\ln\left(1+x^{2}\right)}{x^{2}} = \lim_{u \to 0^{+}}\frac{\ln(1+u)}{u} = 1

Two conditions make the swap legitimate: $u \to 0$ whenever $x \to 0$ , and the standard limit $\frac{\ln(1+u)}{u} \to 1$ holds from either side of $0$ , so using only $u > 0$ costs nothing.

Both sides give the same value

f(-x) = \frac{\ln\left(1+(-x)^{2}\right)}{(-x)^{2}} = \frac{\ln\left(1+x^{2}\right)}{x^{2}} = f(x)

The function is even, so the left-hand and right-hand limits are forced to agree. There is no one-sided behaviour to check here, unlike $\frac{\ln(1+x)}{|x|}$ , whose right-hand limit is $1$ and whose left-hand limit is $-1$ .

The mistakes students make

Two of these are log-law errors, and they are the reason this limit is set as often as it is.

Splitting $\ln\left(1+x^{2}\right)$ into $\ln 1 + \ln x^{2}$ . That is not a log law, and it turns the quotient into $\frac{2\ln|x|}{x^{2}}$ , which runs to negative infinity.
Pulling the square out of the logarithm as a factor in front, then quoting $\frac{\ln\left(1+x^{2}\right)}{x^{2}} \to 1$ and reporting $2 \times 1 = 2$ . The square sits inside $1 + x^{2}$ , so no factor comes out.
Answering $0$ because the numerator vanishes. So does the denominator, and the form is $\frac{0}{0}$ .

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(1 + x^2)/x^2 as x approaches 0?

It is $1$ .

Is substituting u = x^2 allowed inside a limit?

Yes, provided $u \to 0$ as $x \to 0$ , which it does, and the limit in $u$ exists. Both hold here.

Does the side matter for this one?

No. Since $x^{2}$ is never negative, the expression takes the same values at $x$ and $-x$ , so the two one-sided limits are identical.