Need calculus help please

Im having trouble solving these 3 problems. I keep on getting 0's Please help if you can.

Evaluate the limit, if it exists.

1. lim (x->3) (sqrt (x+1) -2) /(x-3)

2. lim(x->2) (1/2+1/x) / (2+x)

3. lim (t->0) (5/(t - 5) / (t^2+t) )

I think you're probably getting 0/0, which is not the same as 0
0/0 is indeterminate, while 0 is not. When you get 0/0 you need to use other methods, such as L'Hopital's Rule, which is shown in first answer.

If you haven't yet seen this method, you can also manipulate the expression algebraically until you no longer get 0/0.

1.

lim[x→3] (√(x+1) − 2) / (x − 3)
= lim[x→3] (√(x+1) − 2)(√(x+1) + 2) / ((x − 3) (√(x+1) + 2))
= lim[x→3] ((x+1) − 4) / ((x − 3) (√(x+1) + 2))
= lim[x→3] (x − 3) / ((x − 3) (√(x+1) + 2))
= lim[x→3]1 / (√(x+1) + 2)
= 1/(√4 + 2)
= 1/4

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Now the other 2 limits do not give 0/0 so you either gave the wrong expression, or the wrong value for x, or else you just need to calculate it.

2.

lim[x→2] (1/2 + 1/x) / (2 + x) = (1/2 + 1/2) / (2 + 2) = 1/4

Perhaps you meant lim[x→−2], which will give 0/0

lim[x→−2] (1/2 + 1/x) / (2 + x)
= lim[x→−2] (x/(2x) + 2/(2x)) / (2 + x)
= lim[x→−2] ((x + 2) / (2x)) / (2 + x)
= lim[x→−2] 1 / (2x)
= 1/(2*−2)
= −1/4

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3.

lim[t→0] (5/(t - 5) / (t^2+t))
= (5/−5 / 0)
= −1/0
= ±∞ depending on whether you are approaching from the left or right

1. If you multiply and divide by the conjugate term, (√(x+1) + 2), this will allow (x - 3) terms to cancel on top and bottom, leaving you with lim{x→3} 1/(√(x+1) + 2) = 1/4

2. Multiply and divide by 2x to get rid of the complex fraction. lim{x→2} (x + 2) / [2x(2 + x)]. Top and bottom terms cancel again, leaving you with lim{x→2} 1/(2x) = 1/4