Math 162

Class 7

Indeterminate Forms

  • \(\frac{0}{0}\)

  • \(\frac{\infty}{\infty}\)

  • \(0\cdot \infty\)

  • \(\infty - \infty\)

  • \(0^0\)

  • \(\infty^0\)

  • \(1^\infty\)

l’Hospital’s Rule

If \(\displaystyle\lim_{x\to a}f(x) = \lim_{x\to a}g(x) = 0\) and if \[\lim_{x\to a}\frac{f'(x)}{g'(x)}\] exists, then \[\lim_{x\to a}\frac{f(x)}{g(x)} = \lim_{x\to a}\frac{f'(x)}{g'(x)}.\]

The same rule works if \(\displaystyle\lim_{x\to a}f(x)\) and \(\lim_{x\to a}g(x) = 0\) are both infinite.

\(\displaystyle \lim_{x \to 0^+ } x \ln x\)

\(\displaystyle \lim_{x \to \infty } x \exp x\)

\(\displaystyle \lim_{x \to -\infty } x \exp x\)

\(\displaystyle \lim_{x \to {\frac{\pi}{2}}^- } \left(\sec x - \tan x\right)\)

\(\displaystyle \lim_{x \to 0^+ } x^{\sin x}\)

\(\displaystyle \lim_{x \to \infty } x^{1/x}\)

\(\displaystyle \lim_{x \to 0^+ } (\cos x)^{1/x^2}\)

Integration by Parts

Given: two functions, differentiable on an interval \((a,b)\), continuous on \([a,b]\).

u v x a b (0 , 0) ff ( a )f ( b )u = f ( x ) gg ( a )g ( b )v = g ( x ) ( f ( x ) ,g ( x ))v ( f ( x )) = g ( x )u ( g ( x )) = f ( x ) f ( b ) f ( a ) v ( u ) du f ( b ) f ( a ) v ( u ) du g ( b ) g ( a ) u ( v ) dv g ( b ) g ( a ) u ( v ) dv + f ( a ) g ( a ) f ( a ) g ( a ) + f ( b ) g ( b )f ( b ) g ( b )= u = f ( x ) du = f ( x ) dx b a v ( f ( x )) f ( x ) dx v = g ( x ) dv = g ( x ) dx b a u ( g ( x )) g ( x ) dx b a g ( x ) f ( x ) dx b a f ( x ) g ( x ) dx

\[{\color{blue}\int_a^b g(x)f'(x)\;dx} + {\color{red}\int_a^b f(x)g'(x)\;dx} + {\color{green}f(a)g(a)} = f(b)g(b)\]

\[ {\color{red}\int_a^b f(x)g'(x)\;dx} = f(b)g(b) - {\color{green}f(a)g(a)} - {\color{blue}\int_a^b g(x)f'(x)\;dx}\]

\[ \int_a^b f(x)g'(x)\;dx = \Biggl.\left(f(x)g(x)\right)\Biggr\rvert_a^b - \int_a^b g(x)f'(x)\;dx\]

Integration by Parts

Let \(f\) and \(g\) be two functions continuous on \([a,b]\) and differentiable on \((a,b)\). Then

\[\int_a^b f(x)g'(x)\;dx = \Biggl.\left(f(x)g(x)\right)\Biggr\rvert_a^b - \int_a^b g(x)f'(x)\;dx\]

This could be useful if \(f'(x)\) is simpler than \(f(x)\) and/or \(g(x)\) is simpler than \(g'(x)\).

\(\displaystyle \int_0^1 x\exp(x)\;dx\)

\(\displaystyle \int_1^2 x\ln x\;dx\)

\(\displaystyle \int_0^{1} \arctan x\;dx\)

Antiderivative Version

\[\int f(x)g'(x)\;dx = f(x)g(x) - \int g(x)f'(x)\;dx\]

Connection to product rule:

\[\phantom{\int} f'(x)g(x)\phantom{\;dx} + \phantom{\int} f(x)g'(x)\phantom{\;dx} = \phantom{\int}\left(f(x)g(x)\right)'\phantom{\;dx}\]

\(\displaystyle \int \ln x\;dx\)

\(\displaystyle \int x^2\cos x\;dx\)

\(\displaystyle \int \cos x \exp(x)\;dx\)