Evaluated at \(\displaystyle x = 1\), the numerator and the denominator are both equal to 0, as shown below:

\(\displaystyle 8^{x}- 8 = 8^{1}- 8 = 8 - 8 = 0\)

\(\displaystyle 2^{x}- 2 = 2^{1}- 2 = 2 - 2 = 0\)

So a straightforward substitution will not work. L'Hospital's rule will work here, but an easier way is to note that

\(\displaystyle 8 = 2 ^{3}\) and \(\displaystyle 8^{x} = \left ( 2 ^{3} \right )^{x} = 2 ^{3x} = \left ( 2 ^{x} \right )^{3}\).

So the expression can be rewritten - and solved - as follows:

\(\displaystyle \lim_{x\rightarrow 1} \frac{8^{x}- 8}{2^{x}- 2}\)

\(\displaystyle = \lim_{x\rightarrow 1} \frac{\left (2^{x} \right )^{3}- 2^{3}}{2^{x}- 2}\)

\(\displaystyle = \lim_{x\rightarrow 1} \frac{\left (2^{x}- 2 \right) \left [ \left ( 2^{x} \right ) ^{2} +2 \cdot 2^{x} + 2 ^{2} \right ] }{2^{x}- 2}\)

\(\displaystyle = \lim_{x\rightarrow 1} \left [ \left ( 2^{x} \right ) ^{2} +2 \cdot 2^{x} + 2 ^{2} \right ]\)

\(\displaystyle = \left [ \left ( 2^{1} \right ) ^{2} +2 \cdot 2^{1} + 2 ^{2} \right ] = 2^{2} + 2 \cdot 2 + 2^{2} = 4 + 4 + 4 = 12\)

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 \(\displaystyle (1)(1)(1)\left(\frac{2}{5}\right) = \left(\frac{2}{5} \right )\).

\(\displaystyle \lim_{x\rightarrow \frac{\pi}{2}^{-} } \cos x = \cos \frac{\pi}{2} = 0\)

and 

\(\displaystyle \lim_{x\rightarrow \frac{\pi}{2}^{-} } \tan x = \infty\),

so we cannot solve this by substituting.

However, we can rewrite the expression:

\(\displaystyle \lim_{x\rightarrow \frac{\pi}{2}^{-} } \cos x \cdot \frac{\sin x}{\cos x}\)

\(\displaystyle = \lim_{x\rightarrow \frac{\pi}{2}^{-} } \sin x = \sin \frac{\pi}{2} = 1\)

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.