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**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.

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

$\displaystyle \color{red}\mbox{little trigonometry:}$

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

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

$\displaystyle \color{red}\lim_{x\rightarrow-\infty}\frac{5x^2-9x}{5x^3+9}=\frac{x-\frac{9}{5x^2}}{1+\frac{9}{5x^3}}$

$\displaystyle \color{red}\mbox{Take it from here (pay attention to the fact that we factored out}\,\, 5x^3$

$\displaystyle \color{red}\mbox{above both in the denominator and in the numerator}$

$\displaystyle \color{red}\mbox{Do the next exercise in a simmilar way}$

$\displaystyle \color{blue}Tonio$

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!