Evaluating Limits

  • Thread starter Destrio
  • Start date
  • #1
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I have been given a series of limits to evaluate, where I can only know if I got 100% of them correct or not, as opposed to individually. So, I'm not sure which I have made a mistake on, so I will post all the limits and my work. Thanks

A)
lim x/|x|
x→ 0

limit does not exist

B)
lim [(x^2)-1]/(x-1)
x→ −1

=(1-1) / (-1-1) = 0/-2 = 0

c)
lim [(x^2)+4x−5]/[(x^2)+x−2]
x→ 1

= (x+5)(x-1)/(x+2)(x-1) = (x+5)/(x+2) = 6/3 = 2

d)
lim |x|
x→ 0

= 0

e)

lim [(2t^2)−3t−2]/[(t^2)+t−6]
t→ 2

= (2t+1)(t-2) / (t+3)(t-2) = (2t+1)/(t+3) = 5/5 = 1

f)

lim[(x^2)−2x+1]/[(x^2)−1]
x→ 1

= (x-1)(x-1) / (x-1)(x+1) = (x-1)/(x+1) = 0/2 = 0

g)
If f(x)=2x−7 find
lim (f(x+h)−f(x)) / h
h→ 0

=2(x+h)-7 - (2x-7) / h
= 2x + 2h -7 - 2x + 7 / h
= 2h / h = 2

h)
If f(x)=(2(x^2)+3x+5) find
lim (f(h)−f(0) )/ h
h→ 0

=2(h^2) + 3h +5 -5 / h
=h(2h+3) / h
=2h+3 = 3


i)
If f(x)=(−25) / (2x+3)
find

lim [f(1+h)−f(1)] / h
h→ 0

= (25/ 2(1+h) +3) - (25/ 2(1) + 3) /h
= (25/ 5+2h) - (5) / h
= (25/ 5+2h) - (5(5+2h))/(5+2h) /h
= (25-25-10h)/(5+2h) /h
= -10 / 5+2h
= -2

j)
lim [(x^2)+h] / [x+(h^2)]
h→ 0

= (x^2) / x
= x



Thanks
 

Answers and Replies

  • #2
1,341
3
For "i" I got 2 -- Other then that looks good
 
  • #3
EnumaElish
Science Advisor
Homework Helper
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I don't see any mistakes except a sign reversal in part i.
 
  • #4
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oh, that would be the one
why is it positive though

shouldnt it be subracting 25 and 10h?
i must be missing something

thanks
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
41,833
956
You said the function was f(x)=(−25) / (2x+3)".

In your calculation, you dropped the "-" from -25 !!
 
  • #6
1,341
3
f(x)=(−25) / (2x+3)

You solved for

If f(x)=(25) / (2x+3)
 

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