I need calc limit help plz!?

Question

No lipatal rule!

1.) Lim

x --> 3 ( 2 x^3 - 6x^2 +x -3 )/(x-3)

2.) Lim

x --> 0 (( x + h) ^3 - x^3)/h

3.) lim

x --> -5- (x+5)/(absolute value ( x+5))

Answer 1

1) lim (2 x^3 - 6x^2 +x -3)/(x - 3)

x-->3

= lim (x - 3)(2x^2 - 6x +1)/(x - 3)

x-->3

= lim (x-3)/(x-3) * lim (2x^2 - 6x + 1)  .....(the1st factor-->1)

x-->3

= lim (2x^2 - 6x + 1)

x-->3

= 2*3^2 - 6*3 + 1

= 1

2) lim [( x + h) ^3 - x^3]/h . . . (apply identity a^3 - b^3 = ...)

h-->0 . . . . . . . . . . . . . . . . . (find limit as h-->0, not x-->0)

= lim (x + h - x)[(x + h)^2 + (x + h)x + x^2] / h

h-->0

= lim h [(x + h)^2 + (x + h)x + x^2] / h

h-->0

= lim h /h * lim [(x + h)^2 + (x + h)x + x^2] ....(1st factor-->1 as h-->0)

h-->0

= 1*lim [(x + h)^2 + (x + h)x + x^2]

. .h-->0

= x^2 + x^2 + x^2

= 3 x^2

3) lim (x+5) / |x+5|

x -->(-5)-

As x approaches (-5) on the left, x-->(- 5) -, i.e. x < (-5), x + 5 < 0,

|x + 5| = -(x + 5)

lim (x+5) / |x+5|

x -->(-5)-

= lim (x+5)/-(x+5)

x -->(-5)-

= (-1) * lim (x+5)/(x+5)

. . . . . x-->(-5)-

= (-1)*1 =(-1)

Answer:

lim (x+5)/|x+5| = (-1)

x -->(-5)-

Answer 2

That s 2 tough

Answer 3

(Problem 1)

Perform polynomial division (either long or synthetic):

Reference for polynomial long division:

http://www.sosmath.com/algebra/factor/fa...

................ 2·x² + 0·x + 1

...... ________________

x - 3 ) 2·x³ - 6·x² + x - 3

...... -(2·x³ - 6·x²)

....... —————

................ 0·x² + x - 3

....................... -(x - 3)

....................... ———

.............................. 0

Therefore:

lim ( 2·x³ - 6·x² + x - 3 ) / ( x - 3 )

x → 3

= lim 2·x² + 1

x → 3

= 2·(3)² + 1

= 19

Answer:

The limit is 19.

——————————————————————————————————————

(Problem 2)

I think you meant as h approaches 0.

Either use binomial theorem to quickly expand or do so by hand (hopefully you can skip some of these steps, but I'm trying to illustrate the method):

lim [ ( x + h )³ - x³ ] / h

h → 0

= lim [ ( x + h )( x + h )( x + h ) - x³ ] / h

h → 0

= lim [ ( x² + 2·h·x + h² )( x + h ) - x³ ] / h

h → 0

= lim [ ( x² + 2·h·x + h² )(x) + ( x² + 2·h·x + h² )(h) - x³ ] / h

h → 0

= lim [ ( x³ + 2·h·x² + h²·x) + ( h·x² + 2·h²·x + h³ ) - x³ ] / h

h → 0

Cancel out the x³. Combine like terms if you want but you can just scratch through a lot of things at this point:

= lim ( 2·h·x² + h²·x + h·x² + 2·h²·x + h³ )/h

h → 0

Now you can divide tough by h:

= lim ( 2·x² + h·x + x² + 2·h·x + h² )

h → 0

= 2·x² + (0)·x + x² + 2·(0)·x + (0)²

= 2·x² + x²

= 3·x²

Answer:

The limit is 3·x²

——————————————————————————————————————

(Problem 3)

Absolute value does one of two things.

(a) If the contents are positive, it does nothing.

(b) If the contents are negative, it changes the sign, which is the same as multiplying by negative one.

In this case, at x=-5, it is zero. So we look at the direction it is approaching from. The -5⁻ means from the left. So it will be negative there always and the absolute value will always change the sign. So change it to:

lim (x+5) / |x+5|

x → -5⁻

= lim (x+5) / -(x+5)

x → -5⁻

Don't distribute the negative, since we already have the same terms. Just cancel:

= lim (1) / -(1)

x → -5⁻

= 1/-1

= -1

Answer:

The limit is -1.

Answer 4

1.) Lim

x --> 3 ( 2 x^3 - 6x^2 +x -3 )/(x-3)

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

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

= lim x->3 2x^2 + 1

= 2*3^2 + 1

=2*9 + 1

=18 + 1

= 19

2.) Lim

x --> 0 (( x + h) ^3 - x^3)/h

= ((0+h)^3 - 0^3)/h

= h^3/h

= h^2

3.) lim

x-->-5-  (x+5)/(absolute value ( x+5))

Put x = -5-h where h is infinitesimal positive quantity

The limit = (-5-h+5)/(absolute value ( -5-h+5))

= -h/(absolute value(-h))

= -h/h

= -1

Answer 5

1)  Factorise and cancel the (x-3) common factor

Then sub in x = 3

2)  Expand numerator to cancel x^3

factorise out the h and cancel h as it will be a common factor

sure you dont mean h->0?

Lim x-> 0 is h^2

3)  becomes +1 when x>-5

-1 when x<-5

are you taking the limit from below?  i.e x<-5

Answer 6

1) lim x --->3

2x^3 ---> 54

6x^2 ---> 54

x - 3 ----> 0

therefore f(x) ----> (x - 3)/(x - 3)

this is because the first two terms will cancel out as x----> 3

this will always equal 1 except when x = 3 which is undefined. and so the limit will be 1.

To do the others use a similar method by finding what each term tends to in the limit