Directly evaluating the limit will produce an indeterminant answer of $\displaystyle \frac{0}{0}$.

Rewriting the limit in terms of sine and cosine, $\displaystyle \lim_{y \rightarrow 0} \frac{\sin 2y}{\cos 2y} \cdot \frac{1}{\sin 5y}$, we can try to manipulate the function in order to utilize the property $\displaystyle \lim_{x \rightarrow 0} \frac{\sin x}{x}=1$.

Multiplying the function by the arguments of the sine functions, $\displaystyle \lim_{y \rightarrow 0} \frac{\sin 2y}{2y} \cdot \frac{1}{\cos 2y} \cdot \frac{5y}{\sin 5y} \cdot \frac{2y}{5y}$, we can see that the limit will be $\displaystyle (1)(1)(1)\left(\frac{2}{5}\right) = \left(\frac{2}{5} \right )$.

The expression $\displaystyle \frac{sin^5(x)}{x^5}$ can be rewritten as $\displaystyle \frac{1}{x^5}sin^5(x)$.

Recall the Squeeze theorem can be used to solve for the limit.  The sine function has a range from $\displaystyle [0,1]$, which means that the range must be inside this boundary.

$\displaystyle 0\leq sin^5(x) \leq 1$

Multiply the $\displaystyle \frac{1}{x^5}$ term through.

$\displaystyle 0\left(\frac{1}{x^5}\right)\leq \left(\frac{1}{x^5}\right)sin^5(x) \leq 1\left(\frac{1}{x^5}\right)$

Take the limit as $\displaystyle x$ approaches infinity for all terms.

$\displaystyle \lim_{x \to \infty}0\leq \lim_{x \to \infty}\frac{sin^5(x)}{x^5} \leq \lim_{x \to \infty}\frac{1}{x^5}$

$\displaystyle 0\leq \lim_{x \to \infty}\frac{sin^5(x)}{x^5} \leq 0$

Since the left and right ends of this interval are zero, it can be concluded that $\displaystyle \lim_{x \to \infty}\frac{sin^5(x)}{x^5}$ must also approach to zero.

The correct answer is 0.

To determine, $\displaystyle \lim_{x \to 0}-\frac{1}{x^2}$, graph the function $\displaystyle y=-\frac{1}{x^2}$ and notice the direction from the left and right of the curve as it approaches $\displaystyle x=0$.

Both the left and right direction goes to negative infinity.

The answer is:  $\displaystyle -\infty$

If $\displaystyle \small \lim _{x\to a} f(x)$ and $\displaystyle \small \lim _{x\to a} g(x)$, then $\displaystyle \small \lim _{x\to a} f(x)+g(x)$ exists.

This can be proven rigorously using the $\displaystyle \small \epsilon-\delta$ definition of a limit, but it is most likely beyond the scope of your class. 

Isolate the constant in the limit.

$\displaystyle \lim_{x \to \infty}\frac{3x^n}{9n!} = \frac{1}{3}\lim_{x \to \infty}\frac{x^n}{n!}$

The limit property $\displaystyle \lim_{x \to \infty}\frac{x^n}{n!}=0$.

Therefore:

$\displaystyle \frac{1}{3}\lim_{x \to \infty}\frac{x^n}{n!} = \frac{1}{3}(0)=0$

To evaluate $\displaystyle \lim_{x\to \infty}tan^{-1}(x^2)$, notice that the inside term $\displaystyle x^2$ will approach infinity after substitution.  The inverse tangent of a very large number approaches to $\displaystyle \frac{\pi}{2} \textup{ radians}$.

The answer is $\displaystyle \frac{\pi}{2}$.

The first step is to factor out the highest degree term from the polynomial on top and bottom (essentially pulling out 1):

$\displaystyle \lim_{x\rightarrow \infty }\frac{x^2}{x^2}\left(\frac{1+\frac{1}{x^2}}{3}\right)$

which becomes

$\displaystyle \lim_{x\rightarrow \infty }\left(\frac{1+\frac{1}{x^2}}{3}\right)$

Evaluating the limit, we approach $\displaystyle \frac{1}{3}$.

To evaluate the limit, first pull out the largest power term from top and bottom (so we are removing 1, in essence):

$\displaystyle \lim_{x\rightarrow \infty }\frac{x^2(\frac{1}{x})}{x^2(2+\frac{3}{x}+\frac{1}{x^2})}$

which becomes

$\displaystyle \lim_{x\rightarrow \infty }\frac{(\frac{1}{x})}{(2+\frac{3}{x}+\frac{1}{x^2})}$

Plugging in infinity, we find that the numerator approaches zero, which makes the entire limit approach 0.

To evaluate the limit, first pull out the highest power term out of the numerator and denominator (so essentially you are pulling 1):

$\displaystyle \lim_{r\rightarrow \infty }\frac{r^3}{r^3}\cdot \frac{3}{1+r^{-2}+r^{-3}}$

As you can see, the $\displaystyle r^{-2}$ and $\displaystyle r^{-3}$ terms as they approach infinity go to zero. What is left over is $\displaystyle \frac{3}{1}=3$.

To evaluate this limit easily, simply pull out a factor of the highest degree term over the highest degree term (1):

$\displaystyle \lim_{g\rightarrow \infty }\frac{g^2}{g^2}\left(\frac{\frac{1}{g}}{1-\frac{2}{g}+\frac{1}{g^2}}\right)$

As you can see, after the $\displaystyle g^2$ divides to 1, then the denominator becomes 1 and the numerator becomes 0.

The final answer is therefore $\displaystyle 0$.