Important Limits

1. Paul's Online Notes - Calculus

L'Hospital's Rule

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Suppose that we have one of the following cases, $$\underset{x\to a}{lim}\frac{f(x)}{g(x)}=\frac{0}{0}\text{ OR }\underset{x\to a}{lim}\frac{f(x)}{g(x)}=\frac{\pm \mathrm{\infty }}{\pm \mathrm{\infty }}$$ where $a$ can be any real number, infinity or negative infinity. In these cases we have, $$\underset{x\to a}{lim}\frac{f(x)}{g(x)}=\underset{x\to a}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)}$$

So, L'Hospital's Rule tells us that if we have an indeterminate form $0/0$ or $\mathrm{\infty }/\mathrm{\infty }$ all we need to do is differentiate the numerator and differentiate the denominator and then take the limit.

$\underset{x\to 0}{lim}\frac{\sin x}{x}$

So, we have already established that this is a 0/0 indeterminate form so let's just apply L'Hospital's Rule. $$\underset{x\to 0}{lim}\frac{\sin x}{x}=\underset{x\to 0}{lim}\frac{\cos x}{1}=\frac{1}{1}=1$$ # $\underset{x\to +\mathrm{\infty }}{lim}\frac{x}{e^x}$

Since $\underset{x\to \mathrm{\infty }}{lim}\frac{x}{e^x}=\frac{\mathrm{\infty }}{\mathrm{\infty }}$ we use the L'Hopital law to get $$\underset{x\to +\mathrm{\infty }}{lim}\frac{x}{e^x}=\underset{x\to +\mathrm{\infty }}{lim}\frac{\frac{dx}{dx}}{d\frac{e^x}{dx}}=\underset{x\to +\mathrm{\infty }}{lim}\frac{1}{e^x}=0$$

$\underset{x\to -\mathrm{\infty }}{lim}x{e}^x$

At first glance, we can see when $x\to -\mathrm{\infty }$, $x\to -\mathrm{\infty }$ and $e^x\to 0$, thus we have $-\mathrm{\infty }\cdot 0$ which is indeterminate.

Howver, we can first use $t=-x$ to substitude $x$ $$\underset{x\to -\mathrm{\infty }}{lim}x{e}^x=\underset{t\to +\mathrm{\infty }}{lim}-t{e}^{-t}$$ Then $$\underset{t\to +\mathrm{\infty }}{lim}-t{e}^{-t}=-\underset{t\to +\mathrm{\infty }}{lim}\frac{t}{e^t}=-\underset{x\to +\mathrm{\infty }}{lim}\frac{x}{e^x}=0$$

$\underset{x\to \mathrm{\infty }}{lim}\frac{{\mathbf{e}}^x}{x^2}$

Sometimes we will need to apply L'Hospital's Rule more than once.

First $$\underset{x\to \mathrm{\infty }}{lim}\frac{{\mathbf{e}}^x}{x^2}=\underset{x\to \mathrm{\infty }}{lim}\frac{{\mathbf{e}}^x}{2x}.$$

Then $$\underset{x\to \mathrm{\infty }}{lim}\frac{{\mathbf{e}}^x}{x^2}=\underset{x\to \mathrm{\infty }}{lim}\frac{{\mathbf{e}}^x}{2x}=\underset{x\to \mathrm{\infty }}{lim}\frac{{\mathbf{e}}^x}{2}=\mathrm{\infty }.$$

Chain Rule

Suppose that we have two functions $f(x)$ and $g(x)$ and they are both differentiable. 1. If we define $F(x)=(f\circ g)(x)$ then the derivative of $F(x)$ is, $$F^{\prime }(x)=f^{\prime }(g(x))g^{\prime }(x)$$ 2. If we have $y=f(u)$ and $u=g(x)$ then the derivative of $y$ is, $$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$$

Each of these forms have their uses, however we will work mostly with the first form in this class.

To see the proof of the Chain Rule see the Proof of Various Derivative Formulas section of the Extras chapter.

Now, let's go back and use the Chain Rule on the function that we used when we opened this section.

Examples

Use the Chain Rule to differentiate $R(z)=\sqrt{5z-8}$.

We've already identified the two functions that we needed for the composition, but let's write them back down anyway and take their derivatives. $$\begin{array}{rlrl}f(z)& =\sqrt{z}& g(z)& =5z-8\\ f^{\prime }(z)& =\frac{1}{2\sqrt{z}}& g^{\prime }(z)& =5\end{array}$$

So, using the chain rule we get, $$\begin{array}{rl}{R}^{\prime }(z)& =f^{\prime }(g(z))g^{\prime }(z)\\ & =f^{\prime }(5z-8)g^{\prime }(z)\\ & =\frac{1}{2}(5z-8{)}^{-\frac{1}{2}}(5)\\ & =\frac{1}{2\sqrt{5z-8}}(5)\\ & =\frac{5}{2\sqrt{5z-8}}\end{array}$$

And this is what we got using the definition of the derivative.