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# 2.7E: Precise Definition of Limit EXERCISES

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## 2.7: The Precise Definition 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) $$\displaystyle \lim_{t→b}\,g(t)=M$$

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

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.

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

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

$$δ≤0.25$$

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.

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

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

$$δ≤2$$

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

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

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

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

$$\frac{x^2−9}{x−3}$$ 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) $$\displaystyle \lim_{x→0}\,x^4=0$$

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

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 $$x=a$$.

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. $$\displaystyle \lim_{x→2}\,f(x)$$

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

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

5

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

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

8/7

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

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

DNE

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

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

2/3

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

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

−4

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

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

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

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.

$$[2,∞]$$

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

$$c=-1$$

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

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

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

$$δ=\sqrt[3]{ε}$$ [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 $$t=[0,2]$$.
$$0$$ m/sec
232) From the previous exercises, estimate the instantaneous velocity at $$t=2$$ by checking the average velocity within $$t=0.01$$ sec.