L'Hopital's Rule is used to evaluate complicated limits. The rule has you take the derivative of both the numerator and denominator individually to simplify the function. In the given function we take the derivatives the first time and get

\(\displaystyle \lim_{x\rightarrow \infty }\frac{2x}{e^x}\). 

This still cannot be evaluated properly, so we will take the derivative of both the top and bottom individually again. This time we get

\(\displaystyle \lim_{x\rightarrow \infty }\frac{2}{e^x}\).

Now we have only one x, so we can evaluate when x is infinity. Plug in infinity for x and we get

\(\displaystyle \frac{2}{e^\infty }\) and \(\displaystyle e^\infty=\infty\). 

So we can simplify the function by remembering that any number divided by infinity gives you zero.

L'Hopital's Rule is used to evaluate complicated limits. The rule has you take the derivative of both the numerator and denominator individually to simplify the function. In the given function we take the derivatives the first time and get 

\(\displaystyle \lim_{x\rightarrow 0}\frac{4x}{\frac{1}{x}}=\lim_{x\rightarrow 0}4x^2\). 

Since the first set of derivatives eliminates an x term, we can plug in zero for the x term that remains. We do this because the limit approaches zero.

This gives us

\(\displaystyle 4\cdot 0^2=0\).

L'Hopital's Rule is used to evaluate complicated limits. The rule has you take the derivative of both the numerator and denominator individually to simplify the function. In the given function we take the derivatives the first time and get 

\(\displaystyle \lim_{x\rightarrow \infty }\frac{15x^2}{6x}\). 

This still cannot be evaluated properly, so we will take the derivative of both the top and bottom individually again. This time we get

\(\displaystyle \lim_{x\rightarrow \infty }\frac{30x}{6}=5x\).

Now we have only one x, so we can evaluate when x is infinity. Plug in infinity for x and we get

\(\displaystyle 5\cdot \infty=\infty\). 

To calculate the limit, often times we can just plug in the limit value into the expression. However, in this case if we were to do that we get \(\displaystyle \frac00\), which is undefined.

What we can do to fix this is use L'Hopital's rule, which says

\(\displaystyle \lim_{x\rightarrow 2} \frac{f(x)}{g(x)}=\lim_{x\rightarrow 2} \frac{f'(x)}{g'(x)}\).

So, L'Hopital's rule allows us to take the derivative of both the top and the bottom and still obtain the same limit.

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

Plug in \(\displaystyle x=2\) to get an answer of \(\displaystyle 2\).

If we plugged in \(\displaystyle -\infty\) directly, we would get an indeterminate value of \(\displaystyle \frac{\infty}{\infty}\).

We can use L'Hopital's rule to fix this. We take the derivate of the top and bottom and reevaluate the same limit.

\(\displaystyle \lim_{x->-\infty}\frac{e^{-2x}}{x^2}=-\frac{e^{-2x}}{x}\).

We still can't evaluate the limit of the new expression, so we do it one more time.

\(\displaystyle \lim_{x->-\infty}-\frac{e^{-2x}}{x}=\lim_{x->-\infty}2e^{-2x}=\infty\)

Subbing in zero into \(\displaystyle \small \frac{(e^{x^{2}}-ln(1+x)-1)}{cosx-sinx-1}\) will give you \(\displaystyle \small \frac{0}{0}\), so we can try to use L'hopital's Rule to solve.

First, let's find the derivative of the numerator. 

\(\displaystyle \small e^{x^{2}}\) is in the form \(\displaystyle \small a^{u}\), which has the derivative \(\displaystyle \small a^{u}u'lna\), so its derivative is \(\displaystyle \small 2xe^{x^{2}}\). 

\(\displaystyle \small ln(1+x)\) is in the form \(\displaystyle \small lnu\), which has the derivative \(\displaystyle \small u'/u\), so its derivative is \(\displaystyle \small 1/(1+x)\).

The derivative of \(\displaystyle \small 1\) is \(\displaystyle \small 0\) so the derivative of the numerator is \(\displaystyle \small 2xe^{x^{2}}-1/(1+x)\).

In the denominator, the derivative of \(\displaystyle \small cosx\) is \(\displaystyle \small -sinx\), and the derivative of \(\displaystyle \small sinx\) is \(\displaystyle \small cosx\). Thus, the entire denominator's derivative is \(\displaystyle \small -(sinx+cosx)\).

Now we take the 

\(\displaystyle \small \lim_{x\rightarrow 0}\frac{2xe^{x^{2}}-1/(x+1)}{-(sinx+cosx)}\), which gives us \(\displaystyle \small \frac{-1}{-1}=1\). 

When you try to solve the limit using normal methods, you find that the limit approaches zero in the numerator and denominator, resulting in an indeterminate form "0/0". 

In order to evaluate the limit, we must use L'Hopital's Rule, which states that:

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

when an indeterminate form occurs when evaluting the limit.

Next, simply find f'(x) and g'(x) for this limit:

\(\displaystyle 2t, -\cos(2-t)\)

The derivatives were found using the following rules:

\(\displaystyle \frac{d}{dx}(x^n)=nx^{n-1}\), \(\displaystyle \frac{d}{dx}\sin(u)=\cos(u)\frac{du}{dx}\)

Next, using L'Hopital's Rule, evaluate the limit using f'(x) and g'(x):

\(\displaystyle \lim_{t\rightarrow 2}\frac{2t}{-\cos (2-t)}=-4\)

Through direct substitution, we see that the limit becomes

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

which is in indeterminate form.

As such we can use l'Hospital's Rule, which states that if the limit 

\(\displaystyle \lim_{x\rightarrow a}\frac{f(x)}{g(x)}\)

is in indeterminate form, then the limit is equivalent to

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

Taking the derivatives we use the power rule which states 

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

Using the power rule the limit becomes 

\(\displaystyle \lim_{x\rightarrow 0}\frac{2x^2-3x}{3x^2+3x} = \lim_{x\rightarrow 0}\frac{2\cdot2x^{2-1}-1\cdot3x^{1-1}}{2\cdot3x^{2-1}+1\cdot3x^{1-1}}\)

\(\displaystyle = \lim_{x\rightarrow 0}\frac{4x-3}{6x+3}\)

\(\displaystyle =-1\)

As such the limit exists and is

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

Through direct substitution, we see that the limit becomes

\(\displaystyle \lim_{x\rightarrow 0}\frac{\sin(\pi\cdot0)}{\sin(0)}=\lim_{x\rightarrow 0}\frac{0}{0}\)

which is in indeterminate form.

As such we can use l'Hospital's Rule, which states that if the limit 

\(\displaystyle \lim_{x\rightarrow a}\frac{f(x)}{g(x)}\)

is in indeterminate form, then the limit is equivalent to

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

Taking the derivatives we use the trigonometric rule which states 

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \sin(Bx)=B\cos(Bx)\) 

where \(\displaystyle B\) is a constant.

Using l'Hospital's Rule we obtain

\(\displaystyle \lim_{x\rightarrow 0}\frac{\sin(\pi x)}{\sin(x)}=\lim_{x\rightarrow 0}\frac{\pi \cos(\pi x)}{\cos(x)}\)

And through direct substitution we find

\(\displaystyle \lim_{x\rightarrow 0}\frac{\pi \cos(\pi x)}{\cos(x)}=\frac{\pi \cos(\pi \cdot0)}{\cos(0)}\)

\(\displaystyle =\frac{\pi \cdot1}{1}\)

\(\displaystyle =\pi\)

As such the limit exists and is

\(\displaystyle \lim_{x\rightarrow 0}\frac{\sin(\pi x)}{\sin(x)}=\pi\)

By substituting the value of \(\displaystyle x=3\), we will find that this will give us the indeterminate form \(\displaystyle \frac{0}{0}\).  This means that we can use L'Hopital's rule to solve this problem.

\(\displaystyle \lim_{x \to 3} \frac{x^2+3x-18}{x-3}=\frac{3^2+3(3)-18}{x-3} = \frac{18-18}{3-3} = \frac{0}{0}\)

L'Hopital states that we can take the limit of the fraction of the derivative of the numerator over the derivative of the denominator.  L'Hopital's rule can be repeated as long as we have an indeterminate form after every substitution.

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

Take the derivative of the numerator.

\(\displaystyle \frac{d}{dx}(x^2+3x-18) = 2x+3\)

Take the derivative of the numerator.

\(\displaystyle \frac{d}{dx}(x-3) =1\)

Rewrite the limit and use substitution.

\(\displaystyle \lim_{x \to 3}\frac{2x+3}{1} = 2(3)+3 =9\)

The limit is \(\displaystyle 9\).