# Thread: Limits and Continuity Help

1. ## Limits and Continuity Help

Can anyone help me out with these 3 calculus problems?

The first one:
f(x) = [1-cos(x)]/(2 sin^2(x))
lim(x-->0)f(x)
f(x) * x/x = [1-cos(x)]/x * x/(2 sin^2(x))
lim(x-->0)[1-cos(x)]/x = 0
lim(x-->0) x/sin x = 1
lim(x-->0) x/sin x = 1
lim(x-->0) 1/(2 sin(x)) = DNE

My teacher told us that the answer is 1/4, but I can't imagine how she got to this point.

For both the 2nd and 3rd- Find the points of discontinuity and label them as removable or nonremovable. I am looking for confirmation, and possibly correction if I am wrong.
The second one:
lim(x--> neg infinity)(5x^4-9x)/(5x^3+9)
Apply direct substitution to get (inf-(neg inf))/(-inf) = -infinity

The third one:
lim(x-->neg inf)sqrt(9x^2+2)/(4x+3)
Apply direct substitution to get (infinity)/(-infinity) = -infinity

Thanks in advance for everyone's help!

2. Originally Posted by macosxnerd101
Can anyone help me out with these 3 calculus problems?

The first one:
f(x) = [1-cos(x)]/(2 sin^2(x))
lim(x-->0)f(x)
f(x) * x/x = [1-cos(x)]/x * x/(2 sin^2(x))
lim(x-->0)[1-cos(x)]/x = 0
lim(x-->0) x/sin x = 1
lim(x-->0) x/sin x = 1
lim(x-->0) 1/(2 sin(x)) = DNE

My teacher told us that the answer is 1/4, but I can't imagine how she got to this point.

$\color{red}\mbox {You can use L'Hospital's Rule and it follows at once, or else use a}$
$\color{red}\mbox{little trigonometry:}$

$\color{blue}\sin^2x = 1-cos^2x\Longrightarrow\frac{1-cosx}{2sin^2x}=\frac{1}{2(1+cosx)}$

For both the 2nd and 3rd- Find the points of discontinuity and label them as removable or nonremovable. I am looking for confirmation, and possibly correction if I am wrong.
The second one:
lim(x--> neg infinity)(5x^4-9x)/(5x^3+9)
Apply direct substitution to get (inf-(neg inf))/(-inf) = -infinity

$\color{red}\lim_{x\rightarrow-\infty}\frac{5x^2-9x}{5x^3+9}=\frac{x-\frac{9}{5x^2}}{1+\frac{9}{5x^3}}$
$\color{red}\mbox{Take it from here (pay attention to the fact that we factored out}\,\, 5x^3$
$\color{red}\mbox{above both in the denominator and in the numerator}$
$\color{red}\mbox{Do the next exercise in a simmilar way}$

$\color{blue}Tonio$

The third one:
lim(x-->neg inf)sqrt(9x^2+2)/(4x+3)
Apply direct substitution to get (infinity)/(-infinity) = -infinity

Thanks in advance for everyone's help!
..

3. Thanks for your help. I'll give #3 a try:

sqrt(9x^2+2)/(4x+3)
[3x/x + sqrt(2)/x] //take 3x out of the radical
divided by (4x/x + 3/x) =
3/4.

Can you confirm this as the right answer?

4. Originally Posted by macosxnerd101
Thanks for your help. I'll give #3 a try:

sqrt(9x^2+2)/(4x+3)
[3x/x + sqrt(2)/x] //take 3x out of the radical
divided by (4x/x + 3/x) =
3/4.

Can you confirm this as the right answer?

Yup, this is right!

Tonio