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April 20, 2014

April 20, 2014

Posted by **angie** on Thursday, March 15, 2012 at 11:32am.

lim x->0 sin5x/2x

lim x->0 1-cosx/x

lim x->7 |x-7|/x-7

lim x->7 (/x+2)-3/x-7

lim h->0 (2+h)^3-8/h

lim t->0 1/t - 1/t^2+t

- Calculus -
**Reiny**, Thursday, March 15, 2012 at 11:57ama lot of your questions contain ambiguous typing

e.g. the second last one probably says

lim ( (2+h)^3 - 8)/h as h --> 0

You could expand the top, or recognize it as a difference of cubes

recall A^3 - B^3 = (A-B)(A^2 + AB + B^2)

= lim [(2+h - 2)((2+h)^2 + 2(2+h) + 4) )/h , h--->0

= lim (2+h)^2 + 2(2+h) + 4 , h--> 0

= 4 + 4 + 4 = 12

for the first one

I will again assume you meant

lim sin (5x) / (2x) , as x --> 0

recall that lim sinØ/Ø = 1 as Ø -->0

so let's "construct" this pattern

multiply top and bottom by (5/2)

so

lim sin (5x) / (2x) , as x --> 0

= lim (5/2)sin (5x)/( (2x)(5/2)

= (5/2) lim sin (5x) / (5x) , as x--->0

= (5/2)(1) = 5/2

try some of the others now after checking on your typing using brackets.

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