as sin(x) approaches 0 as x approaches 0.
That didn't help me as I still resolve to 0*(-∞) but as you say we need a different approach.
I did try to do" let y = (x+sin(x))^2 " and try to take that approach but I didnt know what to do after as now I am introducing y all of a sudden and don't...
1. Does the limit exist of the following:
lim as x→ 1- ((cos^-1(x))/(1-x))
2. Homework Equations :
3. The Attempt at a Solution :
lim as x→ 1- ((cos^-1(x))/(1-x))
= lim as x→ 1- (cos^-1(x))/ lim as x→ 1-(1-x)
Let y = 1-x
lim as y→0 (cos^-1(1-y)) / lim as y→0 (y)
=...
1. lim as x->o+ ( (sqrt(x+sinx))(lnx))
2. Homework Equations : [PLAIN]http://www.math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/SandS/lHopital/Laws/multiplication_law.gif
3. The Attempt at a Solution :
lim as x->o+ ( (sqrt(x+sinx))(lnx))
= lim as x->o+...
The limit division law says if lim of f(x) and g(x) exists and if lim of g(x) isn't 0 then...
Because in my answer the denominator reaches 0, the limit of entire question DOES NOT EXIST ?
I don't know where you got the u from. This is what I got:
Let y= x- \pi so the problem becomes
\lim_{y\to 0} \frac{tan^{-1}(1/-y)}{y}
now I do see that the denominator goes to 0 but the numerator does not. This means the limit does not exist ?
1. lim as x→+pi ##\frac{tan^-1(1/(x-pi))}{pi-x}##
Homework Equations: lim(x/y)= lim (x) - lim (y)
The Attempt at a Solution:
umm don't know where to go from here...
lim as x→+pi [tan^-1(1/(x-pi))] - lim as x→+pi (pi-x)
1. ln(5x+6)+ ln(x-2)=1
2. Homework Equations : log(xy)= log(x)+ log(y)
3. The Attempt at a Solution :
ln(5x+6)+ ln(x-2)=1
ln((5x+6)(x-2))=1
ln(5(x^2)-14x-12)=1
e^(ln(5(x^2)-14x-12)) = e^1
5(x^2)-4x-12=e
5(x^2)-4x-(e-12)=0
use quadratic formula to find x and I get x=2.1617
I don't think I am...
so ...
$$\lim_{x\rightarrow\infty}\frac{3x^3+\cos{x}}{\sin{x}-x^3}=\lim_{x\rightarrow\infty}\frac{3x^3(1+\frac{\cos{x}}{3x^3})}{-x^3(-\frac{\sin{x}}{x^3}+1)}$$ = -3
I GET IT! @Mandelbroth. your new edit was helpful! VERY HELPFUL
Cheers,