When evaluating the limit using normal methods, we receive the indeterminate form $\displaystyle \frac{0}{0}$. When this occurs, we must use L'Hopital's Rule to solve the limit. The rule states that

$\displaystyle \lim_{x\rightarrow a} \frac{f(x)}{g(x)}=\lim_{x\rightarrow a}\frac{f'(x)}{g'(x)}$.

Using the rule, we get

$\displaystyle \lim_{x\rightarrow 0}\frac{\cos(x)}{3x^2+2x+1}=1$

When evaluating the limit using normal methods, we get the indeterminate form $\displaystyle \frac{\infty}{\infty}$. When this occurs, we must use L'Hopital's Rule to evaluate the limit. The rule states that

$\displaystyle \lim_{x\rightarrow a}\frac{f(x)}{g(x)}=\lim_{x\rightarrow a}\frac{f'(x)}{g'(x)}$.

Using the rule for our limit, we get

$\displaystyle \lim_{x\rightarrow 2^+}\frac{\frac{1}{x-2}}{\frac{2x}{x^2-4}}=\lim_{x\rightarrow 2^+}(\frac{1}{(x-2)}\cdot \frac{(x+2)(x-2)}{2x})=1$

We used the following rules to find the derivatives:

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \ln(u)=\frac{1}{u}\frac{\mathrm{d} u}{\mathrm{d} x}$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Evaluating the limit to begin with gets us $\displaystyle \frac{\infty}{\infty}$, which is undefined. We can solve this problem using L'Hospital's rule. Taking the derivative of the numerator and denominator with respect to n, we get $\displaystyle \frac{3n^2+2}{4n}$. The limit is still undefined. Another application of the rule gets us $\displaystyle \frac{6n}{4}$, which evaluated at $\displaystyle n=\infty$ is in fact $\displaystyle \infty$.

In evaluating the limit to begin with, you get $\displaystyle \frac{\infty }{\infty}$, which is undefined. Applying L'Hospitals Rule, we take the derivative of both the numerator and denominator with respect to n. The first derivative gets us $\displaystyle \frac{4n+3}{10n+1}$, which is still improper. Another application of the rule will get us $\displaystyle \frac{2}{5}$, the correct solution. 

Evaluating the limit to begin with, we get $\displaystyle \frac{\infty}{\infty}$, which is undefined. Using L'Hospital's rule to solve, we take the derivative of the numerator and denominator of the expression. In doing so, we get $\displaystyle \frac{4x}{21x^2}$. Evaluating the new limit, we still get $\displaystyle \frac{\infty}{\infty}$. Another application of L'Hospitals rule gives us $\displaystyle \frac{4}{42x}$. We can now solve the limit, which is $\displaystyle \frac{4}{\infty}=0$

To use L'hospital's rule, evaluate the limit of the numerator of the fraction and the denominator separately. If the result is $\displaystyle \infty / \infty$, $\displaystyle -\infty / \infty$, or $\displaystyle 0/0$, take the derivative of the numerator and the denominator separately, and try to evaluate the limit again.

$\displaystyle \lim_{x \to \infty} \frac{2x^3}{x^4+2x^5}$$\displaystyle \longrightarrow \frac{\infty}{\infty}$

$\displaystyle = \lim_{x \to \infty}\frac{6x^2}{4x^3+10x^4}$ (L'hospital's rule) $\displaystyle \longrightarrow \frac{\infty}{\infty}$

$\displaystyle = \lim_{x \to \infty} \frac{12x}{12x^2+40x^3}$ (L'hospital's rule again) $\displaystyle \longrightarrow \frac{\infty}{\infty}$

$\displaystyle = \lim_{x \to \infty}\frac{12}{24x+120x^2}$ (L'hospital's rule again) $\displaystyle \longrightarrow \frac{12}{\infty}$

$\displaystyle = 0$

To use L'hospital's rule, evaluate the limit of the numerator of the fraction and the denominator separately. If the result is $\displaystyle \infty / \infty$, $\displaystyle -\infty / \infty$, or $\displaystyle 0/0$, take the derivative of the numerator and the denominator separately, and try to evaluate the limit again.

$\displaystyle \lim_{x \to 0} \frac{\tan(x^2)}{x^2}$  $\displaystyle \longrightarrow \frac{0}{0}$

$\displaystyle =\lim_{x \to 0} \frac{2x\sec^{2}(x^2)}{2x}$ (L'hospital's Rule) $\displaystyle \longrightarrow \frac{0}{0}$

$\displaystyle =\lim_{x \to 0} \frac{(2x)(2\sec(x^2)(\sec(x^2)\tan(x^2))+\sec^2(x^2)(2)}{2}$ (L'hospital's Rule. Here the derivative of the numerator involves the Product Rule)

$\displaystyle = \frac{(0)(0)+(1)(2)}{2}=1$

To use L'hospital's rule, evaluate the limit of the numerator of the fraction and the denominator separately. If the result is $\displaystyle \infty / \infty$, $\displaystyle -\infty / \infty$, or $\displaystyle 0/0$, take the derivative of the numerator and the denominator separately, and try to evaluate the limit again.

$\displaystyle \lim_{x \to 0} \frac{2^{-x}-1}{3^{-x}-1}$ $\displaystyle \longrightarrow \frac{2^0-1}{3^0-1}\longrightarrow \frac{0}{0}$

$\displaystyle =\lim_{x \to 0} \frac{-\ln(2)(2^{-x})-0}{-\ln(3)(3^{-x})-0}$ (L'hospital's Rule)

$\displaystyle =\frac{\ln(2)}{\ln(3)}$.

So the answer is $\displaystyle \frac{\ln(2)}{\ln(3)}$.

The first step to computing a limit is direct substitution:

$\displaystyle \lim_{x\rightarrow a} \frac{x^3-a^3}{x-a}=\frac{a^3-a^3}{a-a}=\frac{0}{0}.$

Now, we see that this is in the form of L'Hopital's Rule.  For those problems, we take a derivative of both the numerator and denominator separately.  Remember, $\displaystyle a$ is simply a constant!

$\displaystyle \frac{(x^3-a^3)'}{(x-a)'}=\frac{3x^2}{1}.$

Now, we can go back and plug in the original limit value:

$\displaystyle \lim_{x\rightarrow a}\frac{3x^2}{1}=3a^2.$