Limits for functions - No Calculator?

Question

cant use calculator for test but have to find limit. Will I have to prove the limit on the test do you think or can i get it by making assumptions? I am lost and I have a bunch of functions I cant answer because none of them work with direct substitution like this one - lim X------>7 (x^2-49)/(x^2+5x-14) also the same problem with 15 in the denominator instead of 14.

Thanks ahead of time. for all of functions using direct substitution i either get something like 5/0 or 0/0.

Cant use hopitals shortcut either. Must use defeintion of limit

Answer 1

For the given problem, I get 0 / 70 = 0. Perhaps you mean x --> -7? This would give 0/0.

First step should be to factorise:
(x^2 - 49) / (x^2 + 5x - 14) = [(x - 7) (x + 7)] / [(x + 7) (x - 2)]
= (x - 7) / (x - 2) for x ≠ -7.
Now we can substitute -7 to get -14/-9 = 14/9.

In general, if you get something like 5/0 either the limit does not exist (if the denominator is negative on one side and positive on the other) or is either +∞ or -∞ (if the denominator is the same sign on both sides, and depending on the signs of the numerator and denominator).

If you get something like 0/0, first try to factorise and cancel out common factors until either the numerator or denominator (or both) is not 0. It's only when you're left with 0/0 after this step that you need to look at L'Hopital's rule.

Answer 2

lim(x->7) (x^2-49)/(x^2+5x-14) = lim (x+7)(x-7)/(x+7)(x-2)
= lim (x-7)/(x-2) = 0/9 = 0.

now suppose the denominator had -5x instead of +5x, then the original is 0/0 but becomes (x+7)(x-7)/(x+2)(x-7) = (x+7)/(x+2) = 14/9

Answer 3

from what i see, you must FACTOR both numerator and denominator.

so.. x² - 49 = (x + 7)(x - 7)
x² + 5x - 14 = (x + 7)(x - 2)

so (x² - 49) /(x² + 5x - 14) = (x -7) / (x - 2)

some of the factors get cancelled and substitution will get simpler. hopefully

Answer 4

factorise the numerator and the denominator and cancel off the terms (this will usually get rid of the indeterminate form), then apply the limits

0/70 = 0
5/0 can be expressed as infinity
0/0 is indeterrminate form

if you mean lim x---> (-7) check the following link

Answer 5

(x^2-49)/(x^2+5x-14)
=(x+7)(x-7)/[(x+7)(x-2)]
= (x-7)/(x-2)
= 0, when x approaches 7

Answer 6

lim (x -> - ?) [ ?(5x^2 - 2) / (x + 3) ] permit x = - y. as a result, x -> - ? => y -> ? decrease = lim (y -> ?) [ ?(5y^2 - 2) / (- y + 3) ] = lim (y -> ?) [ ?(5 - 2 / y^2) / (- a million + 3 / y) ] (Dividing the numerator and the denominator by utilising y.) = - ?5 (because of the fact, lim y -> ? 2 / y^2 = 0 and lim y -> ? 3 / y = 0.)