AP Calculus BC Limits and Continuity — Worked Answer Explanations

Unit 1 · 12 questions explained

1. Question 1 · Easy

   Evaluate ${\lim }_{x\to 0}\frac{\sin (3x)}{5x}$.

   • A
     $\frac{1}{5}$

     Why not A: Forgets the factor of 3 from $\sin (3x)$; treats it as $\sin (x)/x\cdot 1/5$.
   • B
     $\frac{3}{5}$Correct
   • C
     $\frac{5}{3}$

     Why not C: Inverts the ratio, writing $5/3$ instead of $3/5$.
   • D
     $0$

     Why not D: Incorrectly substitutes $x=0$ and takes only the numerator value $\sin (0)=0$.
   Explanation

   Rewrite: $\frac{\sin (3x)}{5x}=\frac{3}{5}\cdot \frac{\sin (3x)}{3x}$. As $x\to 0$, let $u=3x\to 0$, so $\frac{\sin (3x)}{3x}=\frac{\sin u}{u}\to 1$. Therefore ${\lim }_{x\to 0}\frac{\sin (3x)}{5x}=\frac{3}{5}\cdot 1=\frac{3}{5}$.

   Key takeaway

   For $\lim_{x\to 0}\sin(kx)/(mx)$, factor out $k/m$ and apply the fundamental trig limit $\lim_{u\to 0}\sin(u)/u = 1$.

2. Question 2 · Easy

   For what value of $c$ is $f(x)=\{\begin{array}{ll}{x}^2+c& x<2\\ 3x-1& x\ge 2\end{array}$ continuous at $x=2$?

   • A
     $c=1$Correct
   • B
     $c=-1$

     Why not B: Computes $4+c=5$ but solves with wrong sign, getting $c=5-4-2=-1$.
   • C
     $c=3$

     Why not C: Sets $c=f(2)=5$ then subtracts only $x=2$ (not $x^2=4$), giving $c=5-2=3$.
   • D
     $c=5$

     Why not D: Sets $c$ equal to $f(2)=3(2)-1=5$ without subtracting the $x^2$ term.
   Explanation

   For continuity at $x=2$, we need ${\lim }_{x\to 2^{-}}f(x)=f(2)$. Left limit: ${\lim }_{x\to 2^{-}}(x^2+c)=4+c$. Right value: $f(2)=3(2)-1=5$. Setting equal: $4+c=5\Rightarrow c=1$.

   Key takeaway

   Continuity at a piecewise boundary: set the left-hand limit equal to the right-hand value and solve for the unknown constant.

3. Question 3 · Easy

   Evaluate ${\lim }_{x\to \infty }\frac{4x^3-2x+7}{6x^3+5x^2-1}$.

   • A
     $0$

     Why not A: Treats this like a case where the numerator degree is less than the denominator.
   • B
     $\frac{4}{5}$

     Why not B: Divides leading numerator coefficient by the $x^2$ coefficient of the denominator instead of the $x^3$ coefficient.
   • C
     $\frac{2}{3}$Correct
   • D
     $\infty$

     Why not D: Incorrectly concludes the limit is infinite because the numerator grows without bound.
   Explanation

   Divide numerator and denominator by $x^3$: $\frac{4-2/x^2+7/x^3}{6+5/x-1/x^3}$. As $x\to \infty$ all fractional terms vanish, giving $\frac{4}{6}=\frac{2}{3}$.

   Key takeaway

   When numerator and denominator have equal degree, the limit at infinity equals the ratio of their leading coefficients.

4. Question 4 · Easy

   The Squeeze Theorem guarantees ${\lim }_{x\to 0}{x}^2\sin \left(\frac{1}{x}\right)=0$ because:

   • A
     $\sin (1/x)\to 1$ as $x\to 0$, so the product has limit $0\cdot 1=0$.

     Why not A: $\sin (1/x)$ oscillates and has no limit as $x\to 0$; the product rule for limits cannot be applied here.
   • B
     $-x^2\le x^2\sin (1/x)\le x^2$ and both bounds approach $0$.Correct
   • C
     L'Hôpital's Rule applies since the expression has $0\cdot \infty$ form.

     Why not C: L'Hôpital applies to quotients in $0/0$ or $\infty /\infty$ form; $x^2\sin (1/x)$ is not in that form here.
   • D
     $x^2\sin (1/x)=0$ for all $x\ne 0$.

     Why not D: $\sin (1/x)$ is generally nonzero; the expression is not identically $0$.
   Explanation

   Since $-1\le \sin (1/x)\le 1$ for all $x\ne 0$, multiplying by $x^2\ge 0$ gives $-x^2\le x^2\sin (1/x)\le x^2$. Both $-x^2$ and $x^2$ approach $0$ as $x\to 0$, so by the Squeeze Theorem, ${\lim }_{x\to 0}{x}^2\sin (1/x)=0$.

   Key takeaway

   Use the Squeeze Theorem when a function oscillates but is bounded above and below by functions that converge to the same limit.

5. Question 5 · Easy

   Evaluate ${\lim }_{x\to 0}\frac{1-\cos x}{x^2}$.

   • A
     $0$

     Why not A: Substitutes $x=0$ directly, getting $0/0$, and interprets numerator's leading behavior as $0$ without further analysis.
   • B
     $\frac{1}{2}$Correct
   • C
     $1$

     Why not C: Confuses this with ${\lim }_{x\to 0}\sin (x)/x=1$.
   • D
     $2$

     Why not D: Inverts the correct result, perhaps misapplying the conjugate multiplication.
   Explanation

   Multiply by the conjugate: $\frac{1-\cos x}{x^2}\cdot \frac{1+\cos x}{1+\cos x}=\frac{{\sin }^{2}x}{x^2(1+\cos x)}={\left(\frac{\sin x}{x}\right)}^2\cdot \frac{1}{1+\cos x}$. As $x\to 0$: $(1{)}^2\cdot \frac{1}{1+1}=\frac{1}{2}$. Alternatively, L'Hôpital twice: 2xsinx​→2cosx​​x=0​=21​.

   Key takeaway

   The standard limit $\lim_{x\to 0}(1-\cos x)/x^2 = 1/2$ follows from conjugate multiplication and $\sin x/x \to 1$.

6. Question 6 · Easy

   The function $f(x)=\frac{x^2-4}{x-2}$ has what type of discontinuity at $x=2$?

   • A
     Vertical asymptote

     Why not A: The factor $(x-2)$ cancels, so no vertical asymptote exists at $x=2$.
   • B
     Jump discontinuity

     Why not B: Both one-sided limits equal $4$, so there is no jump.
   • C
     Removable discontinuityCorrect
   • D
     Essential discontinuity

     Why not D: Essential (infinite) discontinuities occur when a limit is infinite; here the limit is finite.
   Explanation

   Factor the numerator: $f(x)=\frac{(x-2)(x+2)}{x-2}=x+2$ for $x\ne 2$. The limit as $x\to 2$ equals $4$, but $f(2)$ is undefined ($0/0$). Since the limit exists but the function is not defined at $x=2$, this is a removable discontinuity — a hole in the graph at $(2,4)$.

   Key takeaway

   A removable discontinuity (hole) occurs when a common factor cancels, leaving a finite limit at a point where the function is undefined.

7. Question 7 · Easy

   Find all horizontal asymptotes of $f(x)=\frac{3x^2-1}{\sqrt{4x^4+5x^2}}$.

   • A
     $y=\frac{3}{2}$ only

     Why not A: Only checks $x\to +\infty$ and overlooks that for $x\to -\infty$ the numerator is still positive, so the sign does not change.
   • B
     $y=\frac{3}{4}$

     Why not B: Computes $\sqrt{4x^4}/x^2=4$ (missing the square root of 4), giving denominator $4$ instead of $2$.
   • C
     $y=\frac{3}{2}$ and $y=-\frac{3}{2}$

     Why not C: Incorrectly introduces a sign flip for $x\to -\infty$; the denominator $\sqrt{4x^4+5x^2}>0$ always, and the numerator $3x^2-1>0$ for large $|x|$, so the ratio stays positive in both directions.
   • D
     $y=\frac{3}{2}$Correct
   Explanation

   For large $|x|$, numerator $\sim 3x^2$ and $\sqrt{4x^4+5x^2}\sim \sqrt{4x^4}=2x^2$. So $f(x)\to \frac{3x^2}{2x^2}=\frac{3}{2}$ as $x\to \pm \infty$. The denominator is always positive, and the numerator is positive for large $|x|$, so the same horizontal asymptote $y=\frac{3}{2}$ applies in both directions.

   Key takeaway

   Check $x \to +\infty$ and $x \to -\infty$ separately; when the denominator involves $\sqrt{x^4} = x^2$ (always positive), only one horizontal asymptote may result.

8. Question 8 · Medium

   For $p(x)=\frac{\sin x}{x(x-\pi )}$, identify and classify all discontinuities on $(-\infty ,\infty )$.

   • A
     Vertical asymptote at $x=0$; vertical asymptote at $x=\pi$.

     Why not A: At $x=0$, $\sin x/x\to 1$, so the singularity is removable, not a vertical asymptote.
   • B
     Removable at $x=0$; removable at $x=\pi$.Correct
   • C
     Removable at $x=0$; vertical asymptote at $x=\pi$.

     Why not C: At $x=\pi$, $\sin \pi =0$ and $(x-\pi )\to 0$; evaluating the limit of $\sin x/(x-\pi )$ as $u=x-\pi \to 0$ gives $-\sin u/u\to -1$, so the limit at $x=\pi$ is $-1/\pi$ (finite), making it removable.
   • D
     Jump discontinuity at $x=0$; removable at $x=\pi$.

     Why not D: Near $x=0$ the function has equal one-sided limits ($-1/\pi$ for the full function), so there is no jump.
   Explanation

   At $x=0$: ${\lim }_{x\to 0}\frac{\sin x}{x(x-\pi )}=\frac{1}{0-\pi }=-\frac{1}{\pi }$ (finite). Removable discontinuity. At $x=\pi$: Let $u=x-\pi$; $\frac{\sin (\pi +u)}{(\pi +u)\cdot u}=\frac{-\sin u}{(\pi +u)u}\to \frac{-1}{\pi \cdot 1}=-\frac{1}{\pi }$ as $u\to 0$. Limit is $-1/\pi$ (finite). Removable discontinuity. Both discontinuities are removable.

   Key takeaway

   When both numerator and denominator vanish at a point, compute the limit carefully — often the singularity is removable, not a vertical asymptote.

9. Question 9 · Hard

   Determine ${\lim }_{x\to 0^{+}}\frac{\ln x}{\cot x}$.

   • A
     $-\infty$

     Why not A: Claims $\ln x\to -\infty$ dominates $\cot x\to +\infty$, but after L'Hôpital the result is finite.
   • B
     $0$Correct
   • C
     $1$

     Why not C: Incorrectly cancels the $-{\csc }^{2}x$ factor or confuses with a $\sin x/x$ type limit.
   • D
     $+\infty$

     Why not D: Reverses numerator/denominator growth rates; $\cot x\to +\infty$ faster than $|\ln x|\to \infty$.
   Explanation

   As $x\to 0^{+}$: $\ln x\to -\infty$ and $\cot x\to +\infty$, giving $-\infty /+\infty$ form. Apply L'Hôpital: $\frac{(1/x)}{-{\csc }^{2}x}=\frac{-{\sin }^{2}x}{x}$. As $x\to 0^{+}$: $\frac{{\sin }^{2}x}{x}=\sin x\cdot \frac{\sin x}{x}\to 0\cdot 1=0$. So the limit is $0$.

   Key takeaway

   L'Hôpital applies to $\pm\infty/\pm\infty$ forms; after differentiating, simplify with standard trig limits to finish.

10. Question 10 · Hard

    For the function $f(x)=\frac{|x^2-4|}{x^2-4}$, evaluate ${\lim }_{x\to 2^{-}}f(x)$ and ${\lim }_{x\to 2^{+}}f(x)$, then classify the discontinuity at $x=2$.

    • A
      Both limits equal $1$; removable discontinuity.

      Why not A: Near $x=2$, $x^2-4<0$ from the left, so $|x^2-4|/(x^2-4)=-1$, not $+1$.
    • B
      ${\lim }_{x\to 2^{-}}f(x)=-1$, ${\lim }_{x\to 2^{+}}f(x)=1$; jump discontinuity.Correct
    • C
      The limit does not exist and $f$ has an infinite discontinuity.

      Why not C: The denominator $x^2-4\to 0$ but the numerator also $\to 0$; cancellation leaves a bounded $\pm 1$ value, not $\pm \infty$.
    • D
      ${\lim }_{x\to 2^{-}}f(x)=1$, ${\lim }_{x\to 2^{+}}f(x)=-1$; jump discontinuity.

      Why not D: Reverses the sign: for $x$ slightly less than $2$, $x^2<4$ so $x^2-4<0$, giving ratio $-1$.
    Explanation

    For $x$ slightly less than $2$: $x^2<4$, so $x^2-4<0$, and $|x^2-4|=-(x^2-4)$. Thus $f(x)=-(x^2-4)/(x^2-4)=-1$. For $x$ slightly greater than $2$: $x^2>4$, so $x^2-4>0$, and $|x^2-4|=x^2-4$. Thus $f(x)=1$. Since ${\lim }_{x\to 2^{-}}f=-1\ne 1={\lim }_{x\to 2^{+}}f$, there is a jump discontinuity at $x=2$.

    Key takeaway

    Absolute value expressions often create jump discontinuities; analyze the sign of the expression inside $|\cdot|$ from each side separately.

11. Question 11 · Hard

    Let $f$ be continuous on $\mathbb{R}$ with $f(x)=\frac{\sin (x^2)}{x}$ for $x\ne 0$. What must $f(0)$ equal?

    • A
      $0$Correct
    • B
      $1$

      Why not B: Confuses ${\lim }_{x\to 0}\sin (x)/x=1$ with ${\lim }_{x\to 0}\sin (x^2)/x$, forgetting the extra factor of $x$ in $x^2$.
    • C
      $-1$

      Why not C: Incorrectly introduces a negative sign from differentiating; the limit here is not via L'Hôpital.
    • D
      $f(0)$ can be any real number.

      Why not D: Continuity uniquely determines $f(0)$; it must equal the limit of $f(x)$ as $x\to 0$.
    Explanation

    For continuity at $x=0$, we need $f(0)={\lim }_{x\to 0}\frac{\sin (x^2)}{x}$. Write $\frac{\sin (x^2)}{x}=x\cdot \frac{\sin (x^2)}{x^2}$. As $x\to 0$: $x^2\to 0$, so $\frac{\sin (x^2)}{x^2}\to 1$, and $x\to 0$. Therefore ${\lim }_{x\to 0}\frac{\sin (x^2)}{x}=0\cdot 1=0$. So $f(0)=0$.

    Key takeaway

    To extend a function continuously, set $f(a) = \lim_{x\to a}f(x)$; factor cleverly to evaluate the limit.

12. Question 12 · Hard

    Evaluate $\underset{x\to \infty }{\lim }{\left(1+\frac{3}{x}\right)}^x$.

    • A
      $1$

      Why not A: Takes the base to $1$ as $x\to \infty$ and concludes $1^{\infty }=1$, ignoring the indeterminate form.
    • B
      $e^3$Correct
    • C
      $3e$

      Why not C: Moves the $3$ outside the exponential without applying the limit correctly.
    • D
      $\infty$

      Why not D: Confuses with a case where the exponent grows faster than the base approaches 1.
    Explanation

    This is the $1^{\infty }$ indeterminate form. Rewrite using the definition $e={\lim }_{n\to \infty }(1+1/n{)}^n$: ${\left(1+\frac{3}{x}\right)}^x={\left[{\left(1+\frac{1}{x/3}\right)}^{x/3}\right]}^3$. As $x\to \infty$, $x/3\to \infty$, so the inner bracket $\to e$. Thus the limit is $e^3$. Alternatively, take the natural log: $x\ln \left(1+\frac{3}{x}\right)\to x\cdot \frac{3}{x}=3$, so the limit is $e^3$.

    Key takeaway

    For $\lim(1+k/x)^x = e^k$; recognize the generalized exponential limit or use $\ln$ to reduce to L'Hôpital.