Aimathic – Limit notation – Free Math Worksheets

Limit notation

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Difficulty: 1 – 5

52190312

Let $f(x)=\begin{cases}\frac{2x+8}{x+5}, &x<-3\\x^2+5x+c, &x>-3\end{cases}$, where $c$ is real. a) Find $\lim_{x\to-3^-}f(x)$. b) Find the value of $c$ for which $\lim_{x\to-3}f(x)$ exists.

Hints

- Use the formula that applies on the side from which $x$ approaches $-3$. - A two-sided limit exists only when the left- and right-hand limits are equal.

Solution

1. For the left-hand limit, use the first formula: $\lim_{x\to-3^-}\frac{2x+8}{x+5}=\frac{2}{2}=1$. 2. For the right-hand limit, use the second formula: $\lim_{x\to-3^+}(x^2+5x+c)=9-15+c=c-6$. 3. The two-sided limit exists when the one-sided limits are equal: $1=c-6$. Thus, $c=7$.

Answer

a) $1$ b) $c=7$

52614012

A hypothetical savings account earns $100\%$ annual interest. If the year is divided into $n$ equal compounding periods, an initial balance of $\$100.00$ grows after one year to $K_n=100\left(1+\frac{1}{n}\right)^n$ dollars. 1. Find the ending balance for annual compounding $(n=1)$, monthly compounding $(n=12)$, and daily compounding $(n=365)$. Round to the nearest cent. 2. Find $\lim_{n\to\infty}K_n$, representing continuous compounding. 3. Find the difference between the daily-compounding balance and the continuous-compounding balance.

Hints

- Substitute each value of $n$ into the model. - Recall the limit that defines $e$. - Keep extra decimal places until the final subtraction.

Solution

1. $K_1=100(2)=\$200.00$. 2. $K_{12}=100\left(1+\frac{1}{12}\right)^{12}\approx\$261.30$. 3. $K_{365}=100\left(1+\frac{1}{365}\right)^{365}\approx\$271.46$. 4. Since $\left(1+\frac{1}{n}\right)^n\to e$, $K_n\to100e\approx\$271.83$. 5. The difference is $100e-K_{365}\approx\$0.37$.

Answer

1. $\$200.00$, $\$261.30$, and $\$271.46$ 2. $100e\approx\$271.83$ 3. Approximately $\$0.37$

52190412

Let $g(x)=\begin{cases}4-x^2, &x<-1\\\frac{6}{x-1}+6, &-1<x<2\\2x+a, &x>2\end{cases}$, where $a$ is real. a) Show that $\lim_{x\to-1}g(x)$ exists, and find it. b) Analyze $\lim_{x\to2}g(x)$ in terms of $a$. For which value of $a$ does the limit exist?

Hints

- Evaluate the one-sided limits using the corresponding formulas. - Set the left- and right-hand limits equal to determine the parameter.

Solution

1. At $x=-1$, the left-hand limit is $4-(-1)^2=3$. 2. The right-hand limit is $\frac{6}{-1-1}+6=-3+6=3$. Since the one-sided limits agree, the limit exists and equals $3$. 3. At $x=2$, the left-hand limit is $\frac{6}{2-1}+6=12$. 4. The right-hand limit is $2(2)+a=4+a$. The two-sided limit exists when $12=4+a$, so $a=8$. For any other value of $a$, the one-sided limits differ.

Answer

a) The limit exists and equals $3$. b) The left-hand limit is $12$, and the right-hand limit is $4+a$. The limit exists only when $a=8$, in which case it equals $12$.

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