L'Hopital's Rule (LHR)

So there’s weird math notation of $\frac{\infty }{\infty }$ or $\frac{0}{0}$ or $0\cdot \infty$ which doesn’t mean what it says. Instead, because we’re dealing with limits, it means “a number which is growing without bound” or “number which is shrinking to zero”. L’Hopital’s rule allows us to resolve these.

Application for limits approaching 0

Let ${\lim }_{x\to c}f(x)=0$ and ${\lim }_{x\to c}g(x)=0$, where f and g are differentiable functions on an open interval $I$ containing $c$, and $g\prime (x)\ne 0$ on $I$ except possibly at $c$. Then:

$$\begin{array}{r}\underset{x\to c}{\lim }\frac{f(x)}{g(x)}=\underset{x\to c}{\lim }\frac{f^{\prime }(x)}{g^{\prime }(x)}\end{array}$$

If the result of applying it still turns into something like $\frac{0}{0}$, you can apply it again as needed.

Application for limits approaching $\infty$

Let ${\lim }_{x\to a}f(x)=\pm \infty$ and ${\lim }_{x\to a}g(x)=\pm \infty$ where f and g are differentiatlbe on an open intervial $I$ containing $a$. Then:

$$\begin{array}{r}\underset{x\to a}{\lim }\frac{f(x)}{g(x)}=\underset{x\to a}{\lim }\frac{f^{\prime }(x)}{g^{\prime }(x)}\end{array}$$

Let f ang g be differentiable functions on the open interval (a, $\infty$), for some value a where $g^{\prime }(x)\ne 0$ on (a, $\infty$) and ${\lim }_{x\to \infty }\frac{f(x)}{g(x)}$ returns either $\frac{0}{0}$ or $\frac{\infty }{\infty }$. Then,

$$\begin{array}{r}\underset{x\to \infty }{\lim }\frac{f(x)}{g(x)}=\underset{x\to \infty }{\lim }\frac{f^{\prime }(x)}{g^{\prime }(x)}\end{array}$$

The same thing works for $-\infty$.

Indeterminite forms (0 * inf)

So the rule only works with fractions. If you get something like $0\cdot \infty$ or $\infty -\infty$, we need to do algebra to get it into the right form.

$$\begin{array}{r}\underset{x\to 0+}{\lim }x\cdot e^{\frac{1}{x}}\end{array}$$

For this limit, x trends down to zero, $e^{\frac{1}{x}}$ trends to $\infty$, so we end up with $0\cdot \infty$.

So we can represent this as a fraction by dividing one side by the inverse of the other:

$$\begin{array}{r}x\cdot e^{\frac{1}{x}}=\frac{e^{\frac{1}{x}}}{\frac{1}{x}}\end{array}$$

To prove this works (because I’ve forgotten lots of algebra), we can do:

$$\begin{array}{r}3\cdot 4=\frac{4}{\frac{1}{3}}\end{array}$$

Plugging into a calculator.. it does equal 12.

Indetermine forms ($0^0$ or $1^{\infty }$)

This means we probably have to use $\ln$.

If ${\lim }_{x\to c}\ln (f(x))=L$, then ${\lim }_{x\to c}f(x)={\lim }_{x\to c}{e}^{ln(f(x))}=e^L$

Example 1

$$\begin{array}{r}\underset{x\to \infty }{\lim }(1+\frac{1}{x}{)}^x\\ \\ x\to \infty \\ \\ 1+\frac{1}{x}\to 1\\ \\ 1^{\infty }\\ \end{array}$$

so $f(x)=1+\frac{1}{x}$

$$\begin{array}{rl}\underset{x\to \infty }{\lim }ln(f(x))& =\underset{x\to \infty }{\lim }ln(1+\frac{1}{x}{)}^x\\ & \\ & =\underset{x\to \infty }{\lim }xln(1+\frac{1}{x})\\ & \\ & =\underset{x\to \infty }{\lim }\frac{ln(1+\frac{1}{x})}{\frac{1}{x}}\end{array}$$

So the transition from 1->2 was due to logarithms property of exponents. Then we were able to do 2->3 b/c of algebra.

From there, we have 0->0, which we can apply LHR with.

$$\begin{array}{rl}& =\underset{x\to \infty }{\lim }\frac{ln(1+\frac{1}{x})}{\frac{1}{x}}\\ & \\ & {=}^{LHR}\underset{x\to \infty }{\lim }\frac{\frac{1}{1+\frac{1}{x}}\cdot -\frac{1}{x^2}}{-\frac{1}{x^2}}\\ & \\ & {=}^{\text{cancel out}}\underset{x\to \infty }{\lim }\frac{1}{1+\frac{1}{x}}\\ & \\ & =1\end{array}$$

Alright, so 2 looks like this b/c we have to do chain rule on the top. The derivative of 1+1/x is -1/x^2 because we need to do -1 * x^(-1-1) (b/c n*x^n-1 for derivative on exponents).

Oof. Okay, so now we go back to our original formula:

$$\begin{array}{r}\underset{x\to \infty }{\lim }ln(f(x));\underset{x\to \infty }{\lim }f(x)=\underset{x\to \infty }{\lim }{e}^{ln(f(x))}=e^L\\ \\ ln(f(x))=1\\ \\ e^1\\ \\ e\end{array}$$

Example 2

$$\begin{array}{r}\underset{x\to 0+}{\lim }{x}^x\\ \\ x\to 0\text{, so }{0}^0\end{array}$$

Slap a $\ln$ on it, then differentiate the insides, the do some algebra hoping to get a fraction.

$$\begin{array}{r}\underset{x\to 0+}{\lim }{x}^x\\ \\ \underset{x\to 0+}{\lim }\ln (x^x)\\ \\ \underset{x\to 0+}{\lim }x\ln (x)\\ \\ \underset{x\to 0+}{\lim }\frac{\ln (x)}{\frac{1}{x}}\\ \\ \underset{x\to 0+}{\lim }\frac{\frac{1}{x}}{-\frac{1}{x^2}}\\ \\ \underset{x\to 0+}{\lim }\frac{x^{-1}}{-x^{-2}}\\ \\ [\text{LHR}]\underset{x\to 0+}{\lim }\frac{x^0}{-x^{-1}}\\ \\ =\frac{1}{-\frac{1}{x}}\\ \\ =-x\\ \\ [\text{LHR}]\underset{x\to 0+}{\lim }-x\\ \\ 0\\ \\ e^0=1.\end{array}$$

Examples

Example 1 (zeros)

$$\begin{array}{r}\underset{x\to 0}{\lim }\frac{sin(x)}{x}\\ \\ \underset{x\to 0}{\lim }\frac{cos(x)}{1}\\ \\ \frac{cos(0)}{1}=1\end{array}$$

Example 2

$$\begin{array}{rl}\underset{x\to 1}{\lim }\frac{\sqrt{x+3}-2}{1-x}& \\ & \\ \underset{x\to 1}{\lim }\frac{\frac{1}{2}(x+3{)}^{-\frac{1}{2}}}{-1}& \\ & \\ & =-\frac{1}{2}\cdot (4{)}^{-\frac{1}{2}}\\ & \\ & =-\frac{1}{2}\cdot \frac{1}{2}\\ & \\ & =-\frac{1}{4}\end{array}$$

Example 3 (infinity)

$$\begin{array}{r}\underset{x\to \infty }{\lim }\frac{3x^2-100x+2}{4x^2+5x-1000}\\ \\ \underset{x\to \infty }{\lim }\frac{6x-100}{8x+5}\\ \\ \frac{6}{8}=\frac{3}{4}\end{array}$$

Example 4 (indeterminite forms)

0 * 0

0 * 0 doesn’t need another intermediate form. It results in 0 (doesn’t need LHR b/c it’s not a differentiatable fraction)

$$\begin{array}{r}\underset{x\to 0-}{\lim }x\cdot e^{\frac{1}{x}}\\ \\ x\to 0\\ \\ e^{\frac{1}{x}}\to 0\\ \\ 0\cdot 0=0\end{array}$$

inf - inf

See: logarithms for relevant rules here

$$\begin{array}{r}\underset{x\to \infty }{\lim }\ln (x+1)-\ln (x)\\ \\ \ln (x)\to \infty \\ \\ \ln (x+1)\to \infty \\ \\ \underset{x\to \infty }{\lim }\ln (\frac{x+1}{x})\\ \\ \frac{x+1}{x}{=}^{LHR}\frac{1}{1}=1\\ \\ \ln (1)=0\end{array}$$

inf - inf

We have to factor out x^2 to get a divisor.

$$\begin{array}{r}\underset{x\to \infty }{\lim }{x}^2-e^x\\ \\ \underset{x\to \infty }{\lim }{x}^2(1-\frac{e^x}{x^2})\\ \\ \frac{e^x}{x^2}{=}^{LHR}\frac{e^x}{2x}{=}^{LHR}\frac{e^x}{2}{=}^{LHR}\infty \\ \\ \underset{x\to \infty }{\lim }{x}^2(1-\infty )\\ \\ \underset{x\to \infty }{\lim }\infty \cdot \infty \\ \\ \infty \end{array}$$

Homework

6.7 9-40, 43, 48-54

27

$$\begin{array}{rl}\underset{x\to -2}{\lim }& \frac{x^3+4x^2+4x}{x^3+7x^2+16x+12}\\ & \\ \underset{x\to -2}{\lim }& \frac{3x^2+6x+4}{3x^2+14x+16}\\ & \\ \underset{x\to -2}{\lim }& \frac{6x+6}{6x+14}\\ & \\ \underset{x\to -2}{\lim }& \frac{6}{6}\\ & \\ & 1\end{array}$$

Quiz Seven

Attempt 1

Q1

$$\begin{array}{rl}& \underset{x\to 0}{\lim }\frac{cos(x)-1}{x^2}\\ & \\ & \underset{x\to 0}{\lim }\frac{-sin(x)}{2x}\\ & \\ & \underset{x\to 0}{\lim }\frac{-cos(x)}{2}\\ & \\ & =-\frac{1}{2}\end{array}$$

Q2

$$\begin{array}{r}\underset{x->0+}{\lim }(2x{)}^x\\ \\ \underset{x\to 0+}{\lim }(2x)\cdot 2\\ \\ \underset{x\to 0+}{\lim }2\cdot 2\\ \end{array}$$

Q3

$$\begin{array}{r}{\int }_1^{\infty }\frac{1}{x^3}\\ \\ {\int }_1^{\infty }\frac{1}{3x^2}\\ \\ {\int }_1^{\infty }\frac{1}{6x}\\ \\ {\int }_1^{\infty }\frac{1}{6}\\ \end{array}$$