Lim (((4 + h)^2)/h) when x is approaching 0?

Question

Can someone please help me and explain how to solve this, I keep reaching 0 as my answer, but 6/5 is the answer in the back of the book. There is no summer tutor available for this class at my college, so any help would be greatly appreciated. Thanks!

Answer 1

the expression: ((4+h)^2)/h

is (16 + 8h +h*h)/h

when evaluated at zero it is
(16/0 + 8 + 0) ... or infinity because the 16/0 is infinity (undefined)

for all h greater than or equal to zero the following is true
4 <= h+4

If this were substituted into the polynomial then the expression would be
Lim((4^2)/h)

because h is positive the following inequality holds
((4^2)/h) <= (((4+h)^2)/h)

therefore
Lim((4^2)/h) <= Lim (((4+h)^2)/h)

Evaluating as h approaches zero yields
inf <= Lim(((4+h)^2)/h)

The two answers agree, therefore the limit is likely to approach infinity. You can also use a tool like excel, MatLab,or Octave to test the idea by plotting it out near zero and see what it starts converging to. When I plot it the graphical solution agrees with the number. If you are familiar with your scientific graphing calculator you can check it graphically using that tool.

Note: the form is not indeterminate, so the use of L'Hopitals rule is not valid. It ONLY applies to indeterminate forms. (0/0, inf/inf, etc...). If you wanted to find the limit of 1/x as x approaches zero, your result would be infinity, a determinate form. If you used L'Hopitals rule on it you would differentiate the numerator and deonminator and get 0/1 or zero. The answer would be invalid.

Answer 2

The limit as *x* goes to 0 is (4+h)^2 /h. The limit as *h* goes to 0 is infinite.

I don't see any variant of this that would give 6/5 as the answer, even assuming various simply typographical errors.

Answer 3

Some of these answers are worng because you can only use L'hopital's rule when x approches infinity. You cannot use L'hopital is x approches zero.

The answer is NOT 0, 6/5, 8 or infinity.

(4 + [really small +#])^2 / ([really small +#])
= 16 / [really small +#]
= [really big +#]
= + ininity

(4 + [really small -#])^2 / ([really small -#])
= 16 / [really small -#]
= [really big -#]
= - ininity

The correct answer is: The Limit Does Not Exist.

The reason is you have to approch zero from both directions. Approch from the left you get negative infinity. Approch from the right you get positive infinity.

Answer 4

Use L'hopital. You have to differntiate both top and bottom as different functions. by L'hopital.

so f(h) = (4+h)^2
f primed = 2(4+h)

g(h) = h
g primed = 1

now we have lim h--->0 of [2(4 +h)]/1]
=8
not sure why 6/5 is answer... i dont think its right.

Answer 5

The limit does not exist, as you approach 0 from the left it is neg inf and as you approach from the pos the lim is pos inf, there is an assymptote at 0.

Answer 6

This is not very difficult assuming that x and h are the same in your question:

if you plot this you will see that as you approach zero from the -1 side it tends to negative infinity and if you approach zero from the +1 side it tends to positive infinity. (do it in excel =(4+A1)^2/A1 and copy it down in cell a1 you have -1 and in cell a2 -.9 etc)

Therefore there is NO LIMIT.

If there were both sides would be converging on it.

Answer 7

h^2 + 8h + 16
-----------------
h

the above equation as the lim h -> 0.

the answer is infinity. h^2/h goes to infinity by l'Hopital rule. Maybe you're reading the answer section wrong.

Answer 8

The answer would be undefined. I think the answer in the book is wrong. Ask your teacher.

Answer 9

Dudette, where is X???
Please re-post the right question

Answer 10

Please restate the question.