When evaluating the limit using normal methods (substitution), we receive the indeterminate form \(\displaystyle \frac{0}{0}\). When we receive the indeterminate form, we must use L'Hopital's Rule to evaluate the limit. The rule states that

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

Using the formula above for our limit, we get

\(\displaystyle \lim_{x\rightarrow 3}\frac{1}{-\frac{\pi}{6}(\sin(\frac{\pi x}{6}))}=-\frac{6}{\pi}\)

The derivatives were found using the following rules:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)\)

When evaluating the limit using normal methods (substitution), we get the indeterminate form \(\displaystyle \frac{0}{0}\). When this happens, to evaluate the limit we use L'Hopital's Rule, which states that

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

Using the above formula for our limit, we get

\(\displaystyle \lim_{x\rightarrow 1}\frac{-1}{4x}=-\frac{1}{4}\)

The derivatives were found using the following rule:

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

To calculate the limit we first plug the limit value into the numerator and denominator of the expression. When we do this we get \(\displaystyle \frac{0}{0}\), which is undefined. We now use L'Hopital's rule which says that if \(\displaystyle f(x)\) and \(\displaystyle g(x)\) are differentiable and

\(\displaystyle f(c)=g(c)=0\),

then 

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

We are evaluating the limit

\(\displaystyle \lim_{x\to 2} \frac{x^3-2x^2+5x-10}{3x-6}\). 

In this case we have 

\(\displaystyle f(x)=x^3-2x^2+5x-10\) 

and

 \(\displaystyle g(x)=3x-6\).

We differentiate both functions and find

 \(\displaystyle f'(x)=3x^2-4x+5\) 

and 

\(\displaystyle g'(x)=3.\) 

By L'Hopital's rule 

\(\displaystyle \lim_{x\to 2} \frac{x^3-2x^2+5x-10}{3x-6}=\lim_{x\to 2} \frac{3x^2-4x+5}{3}\).

When we plug the limit value of 2 into this expression we get 9/3, which simplifies to 3.

If we plugged in the limit value, \(\displaystyle \infty\), directly we would get the indeterminate value \(\displaystyle \frac{\infty}{\infty}\). We now use L'Hopital's rule which says that if \(\displaystyle f(x)\) and \(\displaystyle g(x)\) are differentiable and

\(\displaystyle f(x)=g(x)=\pm\infty\),

then 

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

The limit we wish to evaluate is 

\(\displaystyle \lim_{x\to\infty} \frac{e^x}{\ln(x)}\),

so in this case

\(\displaystyle f(x)=e^x\)

and

\(\displaystyle g(x)=\ln(x)\).

We calculate the derivatives of both of these functions and find that

\(\displaystyle f'(x)=e^x\)

and

\(\displaystyle g'(x)=\frac{1}{x}\).

Thus

\(\displaystyle \lim_{x\to\infty} \frac{e^x}{\ln(x)} = \lim_{x\to\infty} \frac{e^x}{\frac{1}{x}} = \lim_{x\to\infty} x\cdot e^x\).

When we plug the limit value, \(\displaystyle \infty\), into this expression we get \(\displaystyle \infty\cdot\infty\), which is \(\displaystyle \infty\).

We will show that the limit does not exist by showing that the limits from the left and right are different.

We will start with the limit from the right. Using the product rule we rewrite the limit

\(\displaystyle \lim_{x\to 0^+} \frac{e^x}{x} = \lim_{x\to 0^+} \frac{1}{x} \cdot \lim_{x\to 0^+} {e^x}\).

We know that

 \(\displaystyle \lim_{x\to 0} e^x =1\)

and 

\(\displaystyle \lim_{x\to 0^+} \frac{1}{x} = \infty\)

so 

\(\displaystyle \lim_{x\to 0^+} \frac{e^x}{x} =\infty\).

We calculate the limit from the left in the same way and find 

\(\displaystyle \lim_{x\to 0^-} \frac{e^x}{x} =-\infty\).

Thus the two-sided limit does not exist.

To calculate the limit we first plug the limit value into the numerator and denominator of the expression. When we do this we get \(\displaystyle \frac{0}{0}\), which is undefined. We now use L'Hopital's rule which says that if \(\displaystyle f(x)\) and \(\displaystyle g(x)\) are differentiable and

\(\displaystyle f(c)=g(c)=0\),

then 

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

In this case we are calculating 

\(\displaystyle \lim_{x\to 0} \frac{e^x -1 }{x}\)

so

\(\displaystyle f(x)=e^x-1\)

and

\(\displaystyle g(x)=x\).

We calculate the derivatives and find that 

\(\displaystyle f'(x)=e^x\)

and

\(\displaystyle g'(x)=1\).

Thus

\(\displaystyle \lim_{x\to 0} \frac{e^x -1 }{x} = \lim_{x\to 0} \frac{e^x }{1} = 1\).

The limit does not exist. When we first plug in the limit value, 0, we get 0/0 and indeterminate form. We may try to use L'Hopital's rule. However, we quickly discover that 

\(\displaystyle f(x)=|x|\) 

is not differentiable at x=0, so we cannot use L'Hopital's rule. We then consider the limits from the left and right. We determine that 

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

and

\(\displaystyle \lim_{x\to 0^+}\frac{|x|}{x^2+x}=1\).

Thus the two sided limit does not exist.

To calculate the limit we first plug the limit value into the numerator and denominator of the expression. When we do this we get \(\displaystyle \frac{0}{0}\), which is undefined. We now use L'Hopital's rule which says that if \(\displaystyle f(x)\) and \(\displaystyle g(x)\) are differentiable and

\(\displaystyle f(c)=g(c)=0\),

then 

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

We are trying to evaluate the limit

\(\displaystyle \lim_{x\to\frac{\pi}{2}} \frac{\sin^2{x}-1}{\cos{x}}\),

so we have

\(\displaystyle f(x)=\sin^2{x}-1\)

and

\(\displaystyle g(x)=\cos(x)\).

We differentiate both of these functions and find

\(\displaystyle f'(x)=2\sin{x}\cos{x}\)

and

\(\displaystyle g'(x)=-\sin{x}\).

Using L'Hopital's rule we see

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

Now when we plug the limit value into the expression we get \(\displaystyle \frac{0}{-1}\), so 

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

If we plugged in the limit value, \(\displaystyle \infty\), directly we would get the indeterminate value \(\displaystyle \frac{\infty}{\infty}\). We now use L'Hopital's rule which says that if \(\displaystyle f(x)\) and \(\displaystyle g(x)\) are differentiable and

\(\displaystyle f(x)=g(x)=\pm\infty\),

then 

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

The limit we wish to evaluate is 

\(\displaystyle \lim_{x\to\infty} \frac{3x^3+x}{2x^2+3}\),

so we have

\(\displaystyle f(x)=3x^3+x\)

and

\(\displaystyle g(x)=2x^2+3\).

We evaluate the derivatives of these two functions and find

\(\displaystyle f'(x)=9x^2+1\)

and

\(\displaystyle g'(x)=4x\).

Thus, using L'Hopital's we find

\(\displaystyle \lim_{x\to\infty} \frac{3x^3+x}{2x^2+3} = \lim_{x\to\infty} \frac{9x^2+1}{4x}\).

However, when we plug the limit value into this second expression we still end up with the indeterminate value \(\displaystyle \frac{\infty}{\infty}\). So we use L'Hopital's again. 

\(\displaystyle f''(x)=18x\)

and

\(\displaystyle g''(x)=4\).

Using L'Hopital's rule again we see

\(\displaystyle \lim_{x\to\infty} \frac{3x^3+x}{2x^2+1} = \lim_{x\to\infty} \frac{9x^2+3}{4x}= \lim_{x\to\infty} \frac{18x}{4}\).

Plugging in the limit value now we see that the limit evaluates to \(\displaystyle \infty.\)

To calculate the limit we first plug the limit value into the numerator and denominator of the expression. When we do this we get \(\displaystyle \frac{0}{0}\), which is undefined. We now use L'Hopital's rule which says that if \(\displaystyle f(x)\) and \(\displaystyle g(x)\) are differentiable and

\(\displaystyle f(c)=g(c)=0\),

then 

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

We are trying to evaluate the limit

\(\displaystyle \lim_{x\to 0} \frac{\cos{x}-1}{e^x-1}\)

so we have

\(\displaystyle f(x)=\cos{x}-1\)

and

\(\displaystyle g(x)=e^x-1\).

We calculate the derivatives of these functions and get

\(\displaystyle f'(x)=-\sin{x}\)

and

\(\displaystyle g'(x)=e^x\).

Using L'Hopital's rule we find

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

Now when we plug in the limit value we get \(\displaystyle \frac{0}{1}\), so the limit evaluates to 0.