OMPT limits practice: lesson and 10 problems

Answer: 6

Substituting $x=3$ gives $\frac{0}{0}$, so I factor: $\frac{(x-3)(x+3)}{x-3}=x+3$ for $x\ne 3$. The limit only cares about values near 3, so the answer is $3+3=6$.

Answer: B — $\frac{5}{2}$

Dividing top and bottom by $x^2$ gives $\frac{5-3/x}{2+7/x^2}$. As $x\to \infty$ the small terms vanish, leaving $\frac{5}{2}$. With equal degrees, the limit is always the ratio of leading coefficients.

Answer: 4

I match the argument of the sine with the denominator: $\frac{\sin 4x}{x}=4\cdot \frac{\sin 4x}{4x}$. Since $\frac{\sin u}{u}\to 1$ as $u\to 0$, the limit is $4\cdot 1=4$.

Answer: A — $\underset{x\to 0}{\lim }\frac{|x|}{x}$

For $\frac{|x|}{x}$, the value is $1$ from the right and $-1$ from the left; the one-sided limits disagree, so the limit fails to exist. The others all converge: the squeeze theorem gives 0 for the second, factoring gives 4 for the third, and the standard limit gives 1 for the fourth.

Answer: Step 1

The error is the very first move. L'H\^opital's rule only applies to the indeterminate forms $\frac{0}{0}$ or $\frac{\infty }{\infty }$, but here direct substitution works fine: $\frac{1^2+1}{1+1}=\frac{2}{2}=1$. Steps 2 and 3 execute the (inapplicable) rule correctly, which is why the wrong answer 2 looks so convincing.

Answer: 4

Continuity at $x=2$ demands $a={\lim }_{x\to 2}f(x)$. Factoring, $\frac{(x-2)(x+2)}{x-2}=x+2$ near 2, so the limit is 4 and therefore $a=4$.

Answer: -0.25

Substituting $x=-2$ gives $\frac{4-10+6}{4-4}=\frac{0}{0}$, so I factorise both parts: the top is $(x+2)(x+3)$ and the bottom is $(x-2)(x+2)$. Cancelling $(x+2)$ leaves $\frac{x+3}{x-2}$, and at $x=-2$ that is $\frac{1}{-4}=-0.25$. Check with $x=-2.01$: the original fraction is $\frac{-0.0099}{0.0401}\approx -0.247$, sitting next to $-0.25$.

Answer: B — $-\infty$

Substituting gives $\frac{6}{0}$, a non-zero number over zero, so the graph blows up at $x=3$ and I need the sign from the right. Take $x=3.01$: the top is about 6, the bottom is $3-3.01=-0.01$, positive over negative, so the value is $-602$ and it keeps growing in size as $x$ approaches 3. The one-sided limit is $-\infty$. From the left it would be $+\infty$, which is why the two-sided limit does not exist, but the question asked for the right side only.

Answer: 8

For the limit at 3 to exist at all, the numerator must vanish there, otherwise the fraction blows up: $9+3b-12=0$, so $b=1$. With $b=1$ the top is $x^2+x-12=(x+4)(x-3)$, the $(x-3)$ cancels, and the limit is $3+4=7$. Continuity means $c$ equals that limit, so $c=7$ and $b+c=8$. Check: at $x=3.001$, $f\approx \frac{(3.001{)}^2+3.001-12}{0.001}=7.001$.

Answer: 2

A horizontal asymptote is a finite limit at $+\infty$ or at $-\infty$, so I check both. Dividing top and bottom by $x$ needs care, because $\sqrt{x^2}=|x|$. For $x>0$, $\sqrt{x^2+1}=x\sqrt{1+\frac{1}{x^2}}$, so $f(x)\to \frac{2}{1}=2$. For $x<0$, $\sqrt{x^2+1}=-x\sqrt{1+\frac{1}{x^2}}$, so $f(x)\to \frac{2}{-1}=-2$. Two different limits, two asymptotes: $y=2$ and $y=-2$. Check: $f(1000)\approx 1.999999$ and $f(-1000)\approx -1.999999$.