To evaluate the limit, we factor out a term consisting of the highest power term divided by itself (so we are not changing the original function):

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

The term we factored out becomes 1, and all of the negative exponent terms go to zero as n goes to infinity (they are all fractions with an infinitely large denominator, so they equal zero). What is left is $\displaystyle \lim_{n \to \infty} \frac{0}{1}=0$

The first thing to do is to simply plug in the value that $\displaystyle x$ approaches. When you do this:

$\displaystyle \frac{4^3-7\cdot 4^2+2\cdot 4+40}{4^2-11\cdot 4+28}=\frac{64-112+8+40}{16-44+28}=\frac{0}{0}$

When you get this type of indeterminate form, you can use L'Hopital's Rule to evaluate the limit. This rules allows you to find the limit by taking the derivatives of the numerator and denominator.

The derivatives are:

$\displaystyle \lim_{x \to4 }\frac{\frac{d}{dx}(x^3-7x^2+2x+40)}{\frac{d}{dx}(x^2-11x+28)}=\lim_{x \to4 }\frac{3x^2-14x+2}{2x-11}$

The derivatives were found by using the following rule:

$\displaystyle \frac{d}{dx} x^n=nx^{n-1}$

With the derivatives of the numerator and denominator taken, you can again plug in the number that $\displaystyle x$ approaches:

$\displaystyle \frac{3\cdot 4^2-14\cdot 4+2}{2\cdot 4-11}=\frac{-6}{-3}=2$

The first thing to do is simply plug in the value that $\displaystyle x$ approaches:

$\displaystyle \lim_{x \to-\infty }2x^{-\infty} e^{-\infty}=\infty\cdot0$

The best way to find this limit is by using L'Hopital's Rule, but the indeterminate form that was found would not allow for you to use the rule. Therefore, you must manipulate the original function:

$\displaystyle \lim_{x \to-\infty }2x^2 e^x=\lim_{x \to-\infty }\frac{2x^2}{\frac{1}{e^x}}=\lim_{x \to-\infty }\frac{2x^2}{e^{-x}}$

Now, find the limit:

$\displaystyle \frac{2 (-\infty)^2}{e^{\infty}}=\frac{\infty}{\infty}$

This indeterminate form allows for you to use L'Hopital's Rule. This rule states that a limit can be found by taking the derivatives of the numerator and denominator of a function.

The derivatives are:

$\displaystyle \lim_{x \to-\infty }\frac{\frac{d}{dx}2x^2}{\frac{d}{dx}e^{-x}}=\lim_{x \to-\infty }\frac{4x}{-e^{-x}}$

The rules that were used to find the derivatives are:

$\displaystyle \frac{d}{dx}x^n=nx^{n-1},\frac{d}{dx}e^n=e^n\cdot n'$

Now, the value that $\displaystyle x$ approaches can be plugged in to the new function:

$\displaystyle \frac{4\cdot- \infty}{-e^{\infty}}=\frac{-\infty}{-\infty}=\frac{\infty}{\infty}$

Once again, you have an indeterminate form that allows you to apply L'Hopital's Rule again:

$\displaystyle \lim_{x \to-\infty }\frac{\frac{d}{dx}4x}{\frac{d}{dx}-e^{-x}}=\lim_{x \to-\infty }\frac{4}{e^{-x}}$

Now, the value that $\displaystyle x$ approaches can be plugged in to the new function:

$\displaystyle \lim_{x \to-\infty }\frac{4}{e^{-x}}=\frac{4}{\infty}=0$

$\displaystyle \lim_{x\rightarrow 0^+}f(x)$

is defined as the limit as $\displaystyle x$ approaches $\displaystyle 0$ from the right.

Doing so we evaluate

$\displaystyle f(0.1)=e^{0.1}\approx 1.1$

$\displaystyle f(0.01)=e^{0.01}\approx 1.01$

$\displaystyle f(0.001)=e^{0.001}\approx 1.001$

to find that 

$\displaystyle \lim_{x\rightarrow 0^+}f(x)=1$

Step 1: We plug in $\displaystyle 0$ into the function..

We see that the denominator will give me $\displaystyle 0$, which I cannot have.

Step 2: Use L'Hospital Rule:

Take the derivative of both the numerator and denominator until zero doesn't result in the denominator when x is substitute in for a specific value...

Taking derivative once:

$\displaystyle \frac {12x^2-1}{2x}$

Step 3: Plug in $\displaystyle x=-1$ into the new function:

$\displaystyle \frac {12(-1)^2-1}{2(-1)}=\frac {12-1}{-2}=-\frac {11}{2}$

The first step to solving limits is direct substituting the limit constant into the function.  In this case, direct substitution is the only step.

$\displaystyle \lim_{x\rightarrow 0} \frac{x^2-2x}{x^2-1}=\frac{0^2-2(0)}{0^2-1}=\frac{0}{-1}=0.$

The first step to solving limits is plugging in the value specified in the problem.

$\displaystyle \lim_{x\rightarrow -2} \frac{x^2-4}{x+2} = \frac{-2^2-4}{-2+2}=\frac{0}{0}.$

Unfortunately, in this problem, we have a divide by zero after the substitution.  That is not allowed, so we must seek a different method.  The numerator of this fraction is a difference of squares.  Therefore, we need to factor the numerator, which will cancel the factor in the denominator.  Then, we can direct substitute.  

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

$\displaystyle \lim_{x \rightarrow -2}x-2=-2-2=-4.$

To check a one-sided limit, we need to plug numbers into the function close to the value in question. Let's start with a whole number to the right of our value.  The closest whole number is $\displaystyle 4.$

$\displaystyle \frac{1}{4-3}=\frac{1}{1}=1.$

Now, we need to work our way closer to the value in question. Let's try $\displaystyle 3.5.$

$\displaystyle \frac{1}{3.5-3}=\frac{1}{0.5}=2.$

Now, we are getting a larger value.  That is still inconclusive.  We must try one more closer.  Let's do $\displaystyle 3.25.$

$\displaystyle \frac{1}{3.25-3}=\frac{1}{0.25}=4.$

We are even bigger still!  Therefore, we must be increasing to positive infinity. 

To check a one-sided limit, we need to plug numbers into the function close to the value in question. Let's start with a whole number to the left of our value.  The closest whole number is $\displaystyle 1.$

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

Now, we need to work our way closer to the value in question. Let's try $\displaystyle 1.5.$

$\displaystyle \frac{2}{1.5-2}=\frac{2}{-0.5}=-4.$

Now, we are getting a smaller value.  That is still inconclusive.  We must try one more closer.  Let's do $\displaystyle 1.75.$

$\displaystyle \frac{2}{1.75-2}=\frac{2}{-.25}=-8.$

We are even smaller still!  Therefore, we must be decreasing to negative infinity. 

$\displaystyle \begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&\lim_{x\rightarrow n}\frac{f(x)}{g(x)}=\lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{-45sin(x)^{2}}{26x^{2}}\rightarrow\frac{0}{0}\\&\frac{-90cos(x)sin(x)}{52x}\rightarrow\frac{0}{0}\\&\frac{90sin(x)^{2} - 90cos(x)^{2}}{52}\rightarrow\frac{-90}{52}\\&\lim_{x\rightarrow0+}\frac{-45sin(x)^{2}}{26x^{2}}=-\frac{45}{26}\end{align*}$