I need help with these limit problems?

Question

I need help with these limit problems:
1) If (2x - 1 \leq f(x) \leq x^2 - 2x + 3) for (x \geq 0), find (\lim_{{x 	o 2}} f(x)). Evaluate the limit, if it exists. (If the limit does not exist, please enter DNE.)
2) Evaluate the limit: (\lim_{{t 	o 0}} \left(\frac{3}{{t\sqrt{1+t}} - \frac{3}{t}}ight)).
3) Given that (f(x) = -\frac{3}{{x-4}}), find (f(a)), (f(a+h)), and compute (\frac{{f(a+h) - f(a)}}{h}).
4) Find a number (\delta) such that if (|x - 1| < \delta), then (|4x - 4| < \epsilon), where (\epsilon = 1).

Anonymous · Accepted Answer

I'm not trying to be a butthead here, but I remember how it felt trying to learn this stuff too... you will NEED to know this stuff to move on to even more difficult calculus... I will give hints though...
1) use the squeeze theorem
2) do the algebra, if it turns out that there is a vertical asymptote about t, then your answer is DNE or Infinite limits.
3) plug and play with this one, do your algebra and it will work out...
4) again, use the squeeze theorem.

Anonymous · Answer

1) 2x-1= limit f(x)= 3 x->2 2) The limit is zero. 3) f(a)=-3/(a-4) f(a+h)=-3/(a+h-4) (f(a+h)-f(a))/h= [-3/(a+h-4)+3/(a-4)]/h= 3 4) |4x-4|<1=> 4|x-1|<1=> |x-1|<1/4 Choose delta=1/4

Anonymous · Answer

1) As x approaches 2 , 2x-1 approaches 3
x^2-2x+3 approaches 4-4+3 = 3
3 ≤ f(x) ≤ 3
lim x→2 f(x) = 3
2)
http://www.flickr.com/photos/[email protected]/61749476...
3)
f(x)=3/(x-4)
f(a)=3/(a-4)
f(a+h) = 3/(a+h-4)
f(a+h)-f(a) = 3/(a+h-4) - 3/(a-4)
f(a+h)-f(a) = [3(a-4)-3(a+h-4)]/(a-4)(a+h-4)
f(a+h)-f(a) = [3a-12-3a-2h+12] /(a-4)(a+h-4)
f(a+h)-f(a) = -2h /(a-4)(a+h-4)
(f(a+h)-f(a))/h = -2h/h(a-4)(a+h-4) = -2/(a-4)(a+h-4)
4)
ε = 1.
|4x − 4| < 1
|4(x-1)| < 1
|4| |x-1| < 1
4 |x-1| < 1
|x-1| < 1/4
δ =1/4

Anonymous · Answer

The decrease of (x^2 - 4x +8) as x will improve to infinity is x^2. The x^2 term will improve very rapidly, whilst the x term will grow to be bigger quite slowly, and the consistent term will stay small. As x will improve, you would be able to make the smaller words as small as you like relative to x^2. The decrease of the entire expression is 3x^3/x^2 = 3x. Edit: oops, I neglected the unfavorable sign. As x is going to unfavorable infinity, the expression turns into -3x interior the decrease, or unfavorable infinity. See the link for many valuable innovations on limits. a number of them would desire to look obtrusive, yet your elementary experience can usually be tricked by skill of problems like those.

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I need help with these limit problems?

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