The first step is to always plug in the value of the limit. Doing so we get 

$\displaystyle \frac{\infty}{\infty}$

(Remember $\displaystyle e^\infty=\infty$)

This qualifies to use l'Hôpital's Rule on the original equation. The rule says to take the derivative of both the numerator and denominator, individually. We get

$\displaystyle \frac{e^x}{3x^2}$

This still cannot be evaluated therefore we take the derivative of both the numerator and denominator again, individually. We get

$\displaystyle \frac{e^x}{6x}$

Yet again we cannot evaluate this and we need to apply l'Hôpital's Rule again.

$\displaystyle \frac{e^x}{6}$

Now we can evaluate the equation. Plugging in $\displaystyle \infty$ we get.

 $\displaystyle \frac{e^\infty}{6}=\frac{\infty}{6}=\infty$

Which is our answer.

The first step is to always plug in the value of the limit. Doing so we get 

$\displaystyle \frac{\infty}{\infty}$

(Remember $\displaystyle e^\infty=\infty$)

This qualifies to use l'Hôpital's Rule on the original equation. The rule says to take the derivative of both the numerator and denominator, individually. We get

$\displaystyle \frac{2x+2}{e^x}$

This still cannot be evaluated therefore we take the derivative of both the numerator and denominator again, individually. We get

$\displaystyle \frac{2}{e^x}$

Now we can evaluate the equation. Plugging in $\displaystyle \infty$ we get.

 $\displaystyle \frac{2}{e^\infty}=\frac{2}{\infty}=0$

remember that $\displaystyle \frac{constant}{\infty}=0$

Therefore our answer is 0.

The first step is to always plug in the value of the limit. Doing so we get 

$\displaystyle \frac{0}{0}$

This qualifies to use l'Hôpital's Rule on the original equation. The rule says to take the derivative of both the numerator and denominator, individually. We get

$\displaystyle \frac{cos(x)}{1}$

Now we can evaluate the equation. Plugging in $\displaystyle 0$ we get.

 $\displaystyle \frac{cos(0)}{1}=\frac{1}{1}$

Therefore our answer is 1.

Factor the numerator and cancel out like terms

$\displaystyle \lim_{x\rightarrow 2}\frac{x^2-5x+6}{x-2}=\lim_{x\rightarrow 2}\frac{(x-3)(x-2)}{x-2}$

From here, substitute two in for $\displaystyle x$ and solve.

$\displaystyle =\lim_{x\rightarrow 2}(x-3)=-1$

There are two ways to approach this problem. One way is to consider the rates of growth of the numerator and denominator seperately. Both approach infinity, but at different rates. Looking at the graph provided, its clear that the function in the numerator, $\displaystyle y=e^{x}$ grows to infinity at a much greater rate than the function in the denominator, $\displaystyle y=x^{2}$. Since the numerator is growing faster than the denominator, the overall fraction gets larger and larger as x approaches infinity. Thus the answer is $\displaystyle \infty$.

A second way is to use L'hospital's rule a few times.

$\displaystyle \lim_{x \to \infty} \frac{e^{x}}{x^{2}}=\frac{\infty}{\infty}$

$\displaystyle \lim_{x \to \infty} \frac{\frac{\mathrm{d} }{\mathrm{d} x}(e^{x})}{\frac{\mathrm{d} }{\mathrm{d} x}(x^{2})}\Rightarrow \lim_{x \to \infty} \frac{e^{x}}{2x}\Rightarrow \frac{\infty}{\infty}$

$\displaystyle \lim_{x \to \infty} \frac{\frac{\mathrm{d} }{\mathrm{d} x}(e^{x})}{\frac{\mathrm{d} }{\mathrm{d} x}(2x)} \Rightarrow \lim_{x \to \infty} \frac{e^{x}}{2} \Rightarrow \infty$

Either way, the answer is $\displaystyle \infty$.

As $\displaystyle x$ gets closer and closer to zero, $\displaystyle x^2$ also gets closer and closer to zero, and $\displaystyle \frac{1}{x^2}$ gets larger and larger.

In fact,  $\displaystyle \frac{1}{x^2}$ will become arbitrarily large (i.e $\displaystyle \infty$).

To evaluate this limit, we can simply "plug in" 2 for x.

Doing so, we get 

$\displaystyle x^2 - x +2 = (2)^2-(2)+2=4$.

By plugging in 1 for x, we get 

$\displaystyle \frac{x-1}{x^2-1}=\frac{1-1}{1^2-1}=\frac{0}{0}$.

Unfortunately, $\displaystyle \frac{0}{0}$ is an answer in indeterminate form, which means that we cannot make sense of the answer.

Instead, we can use algebra techniques to simplify $\displaystyle f(x)$ first.

Doing so, we get 

$\displaystyle \frac{x-1}{x^2-1}=\frac{x-1}{(x-1)(x+1)}$.

The (x-1) terms cancel, and we are left with $\displaystyle \frac{1}{x+1}$.

Now, by plugging in 1, we get

$\displaystyle \frac{1}{1+1}=\frac{1}{2}$.

By plugging in 0, we get 

$\displaystyle \frac{\sqrt{0^2+9}-3}{0^2}=\frac{0}{0}$.

This solution is "indeterminate." Instead of directly plugging in 0, we simplify the function first, so we multiply by the conjugate of the numerator:

 $\displaystyle \frac{\sqrt{t^2+9}-3}{t^2}*\frac{\sqrt{t^2+9}+3}{\sqrt{t^2+9}+3}$.

Multiplying the numerators (via the FOIL method) and the denominators, we get 

$\displaystyle \\\frac{(\sqrt{t^2+9})^2+3\sqrt{t^2+9}-3\sqrt{t^2+9}-9}{t^2(\sqrt{t^2+9}+3)}\\\\=\frac{t^2+9-9}{t^2(\sqrt{t^2+9}+3)}\\\\=\frac{t^2}{t^2(\sqrt{t^2+9}+3)}\\\\=\frac{1}{\sqrt{t^2+9}+3}$ 

Plugging in 0, we get 

$\displaystyle \frac{1}{\sqrt{0^2+9}+3}=\frac{1}{6}$

By "plugging in" 2, we get $\displaystyle \frac{0}{0}$, which is indeterminate.

We need to manipulate the function to see if anything cancels out first.

By factoring, we have 

$\displaystyle \lim_{x \to 2}\frac{x(x-2)}{(x+1)(x-2)}=\lim_{x \to 2}\frac{x}{(x+1)}=\frac{2}{3}$.