calculusss?

Question

lim as x-> 0

(sin(x^2+x)) / x

Find the limit and show work, i don't get it.

Answer 1

(sin(x^2+x)) / x
lim as x-->0 = 0/0 which is an indeterminate form

Use L'Hopital's Rule:
lim [(derivative of numerator)/(derivative of denominator)]
lim as x-->0 of (2x+1)cos(x^2+x) / 1

Now lim as x-->0 = (2*0 + 1)cos(0 + 0)
= 1 * 1 = 1

Therefore the limit is 1.

Answer 2

Since the lim as x->0 of sin(u)/u where u is the argument is 1 you want to get the denominator into the same thing as the argument.
Thus you multiply top and bottom by X+1

{(X+1)(sin(x^2+X))}/{(X+1)(x)}
which gives you
(X+1)(sin(x^2+X)/(x^2+X)
which is equal to (X+1)*lim as x->0 of Sin(x^2+X)/(x^2+X)
which is equal to 1 since x=0.

Answer 3

Use L'Hospital's rule
You will get lim = 0/1= 0

Answer 4

1, no L'Hospital's rule
---------
Reason:
lim x->0 (sin(x^2+x)) / x
= lim x->0 (sin(0+x)) / x, neglected higher degree terms in x^2+x
= 1