2.7E: Precise Definition of Limit EXERCISES

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2.7: The Precise Defi nition of a Limit

In the following exercises, write the appropriate $ε−δ$ definition for each of the given statements.

176) $\displaystyle \lim_{x→a}\,f(x)=N$

177) $\displaystyle \lim_{t→b}\,g(t)=M$

Answer:: For every $ε>0$, there exists a $δ>0$, so that if $0<|t−b|<δ$, then $|g(t)−M|<ε$

178) $\displaystyle \lim_{x→c}\,h(x)=L$

179) $\displaystyle \lim_{x→a}\,φ(x)=A$

Answer:: For every $ε>0$, there exists a $δ>0$, so that if $0<|x−a|<δ$, then $|φ(x)−A|<ε$

The following graph of the function f satisfies $\displaystyle \lim_{x→2}f(x)=2$. In the following exercises, determine a value of $δ>0$ that satisfies each statement.

$CNX_Calc_Figure_02_05_204.jpeg$

180) If $0<|x−2|<δ$, then $|f(x)−2|<1$.

181) If $0<|x−2|<δ$, then $|f(x)−2|<0.5$.

Answer:: $δ≤0.25$

The following graph of the function f satisfies $\displaystyle \lim_{x→3}\,f(x)=−1$. In the following exercises, determine a value of $δ>0$ that satisfies each statement.

$CNX_Calc_Figure_02_05_205.jpeg$

182) If $0<|x−3|<δ$, then $|f(x)+1|<1$.

183) If $0<|x−3|<δ$, then $|f(x)+1|<2$.

Answer:: $δ≤2$

In the following exercises, use the precise definition of limit to prove the given limits.

J3.7.1) $\displaystyle \lim_{x→5}\,(2x - 1)=9$

Answer:

Let ε$>0$; choose $δ=\frac{ε}{2}$; assume $0<|x−5|<δ$.

In other words:

$0<|x - 5|<\frac{ε}{2}$,

so $-\frac{ε}{2}<x - 5<\frac{ε}{2}$,

then $-ε<2x - 10<ε$

then |$2x - 10|<ε$,

then |($2x - 1)−9$|<ε

Thus, if $0<|x−5|<δ$, then |($2x - 1)−9$|<ε.

Therefore, by the definition of limit, $\displaystyle \lim_{x→5}\,(2x - 1)=9$.

J3.7.2) $\displaystyle \lim_{x→-3}\,(5x+2)=-13$

J3.7.3) $\displaystyle \lim_{x→-7^-}\,\frac{1}{x+7}= −∞ $

Answer:

Let $M>0$; since this is a limit from the left, we need $-δ<x+7<0$ to lead $\frac{1}{x+7}<-M$ (since the limit is $−∞ $)
Note: since $x<-7, x+7<0$

Proof:

Let $M>0$. Choose $δ=\frac{1}{M}$.

If $-δ<x+7<0$, in other words $-\frac{1}{M}<x+7<0$

then $\frac{1}{M}>-(x+7)>0$ (both $\frac{1}{M}$ and $-(x+7)$ are positive)

then $M<-\frac{1}{x+7}$

then $-M>\frac{1}{x+7}$

so $\frac{1}{x+7}<-M$

Thus, if $-\frac{1}{M}<x+7<0$ then$\frac{1}{x+7}<-M$

Therefore, by the definition of (infinite, from the left) limit, $\displaystyle \lim_{x→-7^-}\,\frac{1}{x+7}= −∞ $.

J3.7.4) $\displaystyle \lim_{x→2^+}\,\frac{1}{x-2}= ∞ $

188) $\displaystyle \lim_{x→2}\,(5x+8)=18$

189) $\displaystyle \lim_{x→3}\,\frac{x^2−9}{x−3}=6$

Answer:: $\frac{x^2−9}{x−3}$ is equivalent to $x + 3$ by factoring, as long as $x$ is not $3$. Since we are looking at the limit as $x→3$, we do not consider $x=3$.
Let $ε>0$, choose $δ=ε$. If $0<|x−3|<ε$, then $|x+3−6|=|x−3|<ε$. Thus, by the definition of limit, $\displaystyle \lim_{x→3}\,\frac{x^2−9}{x−3}=6$.

190) $\displaystyle \lim_{x→2}\,\frac{2x^2−3x−2}{x−2}=5$

191) $\displaystyle \lim_{x→0}\,x^4=0$

Answer:: Let $ε>0$, choose$δ=\sqrt[4]{ε}$ If $0<|x|<\sqrt[4]{ε}$, then $∣x^4∣=x^4<ε$. Thus, by the definition of limit, $\displaystyle\lim_{x→0},x^4=0$.

192) $\displaystyle \lim_{x→2}\,(x^2+2x)=8$

Chapter Review Exercises

True or False. In the following exercises, justify your answer with a proof or a counterexample.

208) A function has to be continuous at $x=a$ if the $\displaystyle \lim_{x→a}\,f(x)$ exists.

209) Evaluate $\displaystyle \lim_{x→0}\,\frac{sinx}{x}$ = ?

Answer:: 1

210) If there is a vertical asymptote at $x=a$ for the function $f(x)$, then f is undefined at the point $x=a$.

211) If $\displaystyle \lim_{x→a}\,f(x)$ does not exist, then f is undefined at the point $x=a$.

Answer:: False. A removable discontinuity is possible.

212) Using the graph, find each limit or explain why the limit does not exist.

a. $\displaystyle \lim_{x→−1}\,f(x)$

b. $\displaystyle \lim_{x→1}\,f(x)$

c. $\displaystyle \lim_{x→0^+}\,f(x)$

d. $\displaystyle \lim_{x→2}\,f(x)$

$CNX_Calc_Figure_02_05_207.jpeg$

In the following exercises, evaluate the limit algebraically or explain why the limit does not exist.

213) $\displaystyle \lim_{x→2}\,\frac{2x^2−3x−2}{x−2}$

Answer:: 5

214) $\displaystyle \lim_{x→0}\,3x^2−2x+4$

215) $\displaystyle \lim_{x→3}\,\frac{x^3−2x^2−1}{3x−2}$

Answer:: 8/7

216) $\displaystyle \lim_{x→π/2}\,\frac{cotx}{cosx}$

217) $\displaystyle \lim_{x→−5}\,\frac{x^2+25}{x+5}$

Answer:: DNE

218) $\displaystyle \lim_{x→2}\,\frac{3x^2−2x−8}{x^2−4}$

219) $\displaystyle \lim_{x→1}\,\frac{x^2−1}{x^3−1}$

Answer:: 2/3

220) $\displaystyle \lim_{x→1}\,\frac{x^2−1}{\sqrt{x}−1}$

221) $\displaystyle \lim_{x→4}\,\frac{4−x}{\sqrt{x}−2}$

Answer:: −4

222) $\displaystyle \lim_{x→4}\,\frac{1}{\sqrt{x}−2}$

In the following exercises, use the squeeze theorem to prove the limit.

223) $\displaystyle \lim_{x→0}\,x^2cos(2πx)=0$

Answer:: Since $−1≤cos(2πx)≤1$, then $−x^2≤x^2cos(2πx)≤x^2$. Since $lim_{x→0}\,x^2=0=lim_{x→0}\,−x^2$, it follows that $lim_{x→0}\,x^2cos(2πx)=0$.

224) $\displaystyle \lim_{x→0}\,x^3sin(\frac{π}{x})=0$

225) Determine the domain such that the function $f(x)=\sqrt{x−2}+xe^x$ is continuous over its domain.

Answer:: $[2,∞]$

In the following exercises, determine the value of c such that the function remains continuous. Draw your resulting function to ensure it is continuous.

226) $f(x)=\begin{cases}x^2+1 & x>c\\2^x & x≤c\end{cases}$

227) $f(x)=\begin{cases}\sqrt{x+1} & x>−1\\x^2+c & x≤−1\end{cases}$

Answer:: $c=-1$

In the following exercises, use the precise definition of limit to prove the limit.

228) $\displaystyle \lim_{x→1}\,(8x+16)=24$

229) $\displaystyle \lim_{x→0}\,x^3=0$

Answer:: $δ=\sqrt[3]{ε}$ [This is just a piece for constructing the proof.]

230) A ball is thrown into the air and the vertical position is given by $x(t)=−4.9t^2+25t+5$. Use the Intermediate Value Theorem to show that the ball must land on the ground sometime between 5 sec and 6 sec after the throw.

231) A particle moving along a line has a displacement according to the function $x(t)=t^2−2t+4$, where x is measured in meters and t is measured in seconds. Find the average velocity over the time period $t=[0,2]$.

Answer:: $0$ m/sec

232) From the previous exercises, estimate the instantaneous velocity at $t=2$ by checking the average velocity within $t=0.01$ sec.