Limit Questions?

Question

Need homework help:

lim(x->1) (1-x)/sqrt(12-3x)-3

lim(x->1) (x^3-1)/(x-1)

lim(x->0) (1-cosx)/(1+sinx)

lim (x->0) (1-cosx)/(sinx)

Don't have to do all of them, just please show me your steps.

Answer 1

One trick to doing limits of functions, is to separate your function into chunks where limits are easy to find.

e.g., the first one

Lim(x->1) [ (1-x)/SQRT(12-3x) - 3]
=
Lim (1-x) / Lim(SQRT(12-3x)) - Lim(3)

Lim(3) = 3 whatever x does.

Lim(1-x) = 0 as x approaches 1.

Lim(SQRT(12-3x)) = SQRT(9) as x approaches 1.

Replacing, we have
Lim(x->1) [ (1-x)/SQRT(12-3x) - 3]
=
0 / 9 - 3 = -3

(Depending where the "-3" belongs)

The second one may be a bit of a problem:
lim(x->1) (x^3-1)/(x-1)
If we try directly, we get
lim(x^3-1) = 0 and
lim(x-1) = 0
0/0 = indeterminate.

However, we can split it differently:

(x^3-1) = (x-1)(x^2+x+1)

Therefore:
lim(x->1) (x^3-1)/(x-1) =
lim(x^2+x+1)*lim[(x-1)/(x-1)] =
lim(x^2+x+1)*lim(1)

lim(x^2+x+1) = 3 as x approaches 1
lim(1) = 1 whatever x does.
3*1 = 3

This is NOT the same as saying that the original function is equal to 3 when x = 1 (after all, the original function continues to have a division by zero at that position). However, it can be completed by giving it the value of the limit at x=1.

Answer 2

I'll do two. You do the others, but let me know if you need help.

lim(x->1) (x^3-1)/(x-1)
Plug in 1 and you get 0/0, an indeterminate form. So you have to simplify:
Difference of cubes for numerator:
lim(x->1) [(x-1)(x^2 + x + 1)]/(x-1)
lim(x->1) (x^2 + x + 1)
= 1^2 + 1 + 1 = 3

lim(x->0) (1-cosx)/(1+sinx)
You can just plug in for this one:
= [1 - cos(0)] / [1 + sin(0)]
= [1 - 1] / [1 + 0]
= 0

Answer 3

lim(x->1) (1-x)/sqrt(12-3x)-3
= lim x->1 (1-x) [sqrt(12-3x) +3]/ [sqrt(12-3x)-3 ] [sqrt(12-3x) +3]
= lim x->1 (1-x) [sqrt(12-3x) +3]/ [ 12-3x-9]
= lim x->1 (1-x) [sqrt(12-3x) +3]/ [ 3-3x]
= lim x->1 (1-x) [sqrt(12-3x) +3]/ 3[ 1-x]
= lim x->1 [sqrt(12-3x) +3]/ 3
= [sqrt(9) +3]/3 = 2


lim(x->1) (x^3-1)/(x-1)
= lim x->1 (x-1)(x^2 +x +1)/ (x-1)
= lim x->1 x^2 +x +1 = 3

Answer 4

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