If we plugged in the integration limit to the expression in the problem we would get \(\displaystyle \frac{\infty}{\infty}\), which is undefined. Here we use L'Hopital's rule, which is shown below.

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

This gives us, 

\(\displaystyle \lim_{x\rightarrow \infty} \frac{5x^2+2x+1}{x^4+2}=\lim_{x\rightarrow \infty} \frac{10x+2}{4x^3}\).

However, even with this simplified limit, we still get \(\displaystyle \frac{\infty}{\infty}\). So what do we do? We do L'Hopital's again!

\(\displaystyle \lim_{x\rightarrow \infty} \frac{10x+2}{4x^3}=\lim_{x\rightarrow \infty} \frac{10}{12x^2}\).

Now if we plug in infitinity, we get 0. 

\(\displaystyle \frac{10}{\infty}=0\)

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 \(\displaystyle \small \frac{-1}{-1}=1\). 

If we plug in 0 into the limit we get \(\displaystyle \frac{0}{0}\), which is indeterminate.

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

\(\displaystyle \lim_{x->0}\frac{sin(x)}{ln(x)}=\lim_{x->0}xcos(x)=0\).

Now if we plug in 0, we get 0, so that is our final limit.

If we try to directly plug in the limit value into the function, we get

\(\displaystyle \small \small \frac{\cos 0-(0-1)^2}{0^3-3\cdot 0}=\frac{1-1}{0-0}=\frac{0}{0}\)

Because the limit is of the form \(\displaystyle \small \frac{0}{0}\), we can apply L'Hopital's rule to "simplify" the limit to

\(\displaystyle \small \small \lim_{x\to 0} \frac{\cos x-(x-1)^2}{x^3-3x}=\lim_{x\to 0} \frac{\sin x-2(x-1)}{3x^2-3}\).

Now if we directly plug in 0 again, we get

\(\displaystyle \small \small \small \lim_{x\to 0} \frac{\cos x-(x-1)^2}{x^3-3x}=\lim_{x\to 0} \frac{\sin x-2(x-1)}{3x^2-3}=\frac{0-2(0-1)}{0-3}=-\frac{2}{3}\).

In order to determine the limit, substitute \(\displaystyle x=0\) to determine whether the expression is indeterminate.  

\(\displaystyle \lim_{x \to 0} \frac{cos(x)-1}{tan(x)} = \frac{0}{0}\)

Use the L'Hopital's rule to simplify.  Take the derivative of the numerator and denominator separately, and reapply the limit.  

\(\displaystyle \lim_{x \to 0} \frac{-sin(x)}{sec^2(x)}= -sin(x)\times \frac{1}{cos(x)}\times \frac{1}{cos(x)}\)

Substitute \(\displaystyle x=0\)

\(\displaystyle -sin(0)\times \frac{1}{cos(0)}\times \frac{1}{cos(0)} =0\)

By substitution, the limit will yield an indeterminate form \(\displaystyle \frac{0}{0}\).  L'Hopital can be used in this scenario.

Take the derivative of the numerator and denominator separately, and then reapply the limit.

\(\displaystyle \lim_{x\to 3}\frac{(x-3)^2}{(3-x)^2}=\lim_{x\to 3}\frac{2(x-3)}{-2(3-x)}=\lim_{x\to 3}-\frac{x-3}{3-x}=\lim_{x\to 3}-\frac{x-3}{-(x-3)}=1\)

The answer is \(\displaystyle 1\).

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):

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

Rewrite the expression.

\(\displaystyle \lim_{n\to \infty} (\frac{n+1}{n})^2= \lim_{n\to \infty} \frac{(n+1)^2}{n^2}\)

By substitutition, we will get the indeterminate form \(\displaystyle \frac{\infty}{\infty}\).

The L'Hopital's rule can be used to solve for the limit.  Write the L'Hopital's rule.

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

Apply this rule twice.

\(\displaystyle \lim_{n\to \infty} \frac{(n+1)^2}{n^2}=\lim_{n\to \infty} \frac{2(n+1)}{2n}=\lim_{n\to \infty} \frac{2}{2}=1\)

When evaluating the limit using normal methods (substitution), you find that the indeterminate form of \(\displaystyle \frac{0}{0}\) is reached. To evaluate the limit, we can use L'Hopital's Rule, which states that:

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

So, we find the derivative of the numerator and denominator, which is

\(\displaystyle 1\) and \(\displaystyle \cos(x)\), respectively.

The derivatives were found using the following rules:

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

When we evaluate this new limit, we find that

\(\displaystyle \lim_{x\rightarrow 0}\frac{1}{cos(x)}=1\).