Practice: Limits and Continuity

(a) $\underset{x\to 3}{lim}{\displaystyle \frac{x^2-9}{x^2-x-6}}$. Substituting $x=3$: the numerator is $9-9=0$ and the denominator is $9-3-6=0$. The form is $\frac{0}{0}$, so the quotient law does not apply, its hypothesis requires the denominator's limit to be nonzero. Both parts vanish at 3, so both carry a factor of $(x-3)$:

The cancellation is legitimate because the limit never evaluates at $x=3$. Now the denominator tends to $5\ne 0$, so the quotient law applies:

Check: at $x=3.001$ the original expression is $1.19996$, at $x=2.999$ it is $1.20004$ ✓. (b) $\underset{x\to 0}{lim}{\displaystyle \frac{\sqrt{x+4}-2}{x}}$. Substitution gives $\frac{2-2}{0}=\frac{0}{0}$ again. Here factoring is unavailable, so rationalise by multiplying above and below by the conjugate $\sqrt{x+4}+2$:

The $x$ cancels, and the remaining expression is continuous at 0:

Check: at $x=0.0001$ the original expression is $0.24999844$, and at $x=-0.0001$ it is $0.25000156$ ✓. (c) The piecewise function at $x=1$. From the left, so $f$ follows $x^2$:

From the right, $x\ge 1$ so $f$ follows $3x$:

Both one-sided limits exist and they differ, so the two-sided limit does not exist. The discontinuity is a jump, of size $3-1=2$. Values confirm it: $f(0.999)=0.998$ and $f(1.001)=3.003$. Redefinition cannot repair it. Continuity at 1 would require $f(1)$ to equal $\underset{x\to 1}{lim}f(x)$, and no such limit exists. No single number equals both 1 and 3. This is the difference from a removable discontinuity, where the limit exists and only the value is missing or misplaced. Note that $f(1)=3$ is already defined; having a value was never the problem. (d) $\underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{3x^2+2x}{x^2-5}}$. Divide numerator and denominator by $x^2$, the highest power present:

Check: $3.3684$ at $x=10$, $3.0215$ at $x=100$, $3.00202$ at $x=1000$, $3.000002$ at $x={10}^6$ ✓. What it says about the graph. The line $y=3$ is a horizontal asymptote: as $x$ grows the curve levels off toward 3 without reaching it. Since the degrees are equal, the limit is the ratio of leading coefficients, and the same computation as $x\to -\mathrm{\infty }$ gives 3 as well, so the asymptote holds in both directions. (e) What is wrong with the reasoning. The value 1 is correct, but the argument is not, and it would produce false answers elsewhere. "$\frac{0}{0}$ is 1 because anything over itself is 1" treats $\frac{0}{0}$ as an arithmetic expression to simplify. It is not a number. The numerator and denominator are two different functions that happen both to tend to 0; nothing says they do so at the same rate, and the rate is what determines the limit. Three limits of the form $0/0$, as $x\to 0$: | Limit | Value |
|---|---|
| ${\displaystyle \frac{x}{x}}$ | $1$ |
| ${\displaystyle \frac{x^2}{x}}$ | $0$ |
| ${\displaystyle \frac{x}{x^2}}$ | no finite limit; grows without bound | All three have the same form and three different answers, which is exactly why the form settles nothing. The reasoning offered would assign 1 to all three. What the form actually signals. It says the quotient law's hypothesis has failed, so the laws cannot be applied as written, and further work is needed, factoring, rationalising, a squeeze, or a theorem such as L'Hôpital's rule. It is a diagnosis of the obstacle, not a result. Why $\frac{\sin x}{x}\to 1$ in fact. No algebraic manipulation clears this one; the standard argument bounds $\cos x\le \frac{\sin x}{x}\le 1$ near 0 by comparing areas in the unit circle and applies the squeeze theorem, since both bounds tend to 1. That the correct value emerged from incorrect reasoning is a coincidence of this particular case.