L'Hôpital's rule review

L'Hôpital's rule helps us evaluate indeterminate limits of the form $\dfrac{0}{0}$start fraction, 0, divided by, 0, end fraction or $\dfrac{\infty}{\infty}$start fraction, infinity, divided by, infinity, end fraction.

In other words, it helps us find $\displaystyle\lim_{x\to c}\dfrac{u(x)}{v(x)}$limit, start subscript, x, \to, c, end subscript, start fraction, u, left parenthesis, x, right parenthesis, divided by, v, left parenthesis, x, right parenthesis, end fraction, where $\displaystyle\lim_{x\to c}u(x)=\lim_{x\to c}v(x)=0$limit, start subscript, x, \to, c, end subscript, u, left parenthesis, x, right parenthesis, equals, limit, start subscript, x, \to, c, end subscript, v, left parenthesis, x, right parenthesis, equals, 0 (or, alternatively, where both limits are $\pm\infty$plus minus, infinity).

The rule essentially says that if the limit $\displaystyle\lim_{x\to c}\dfrac{u'(x)}{v'(x)}$limit, start subscript, x, \to, c, end subscript, start fraction, u, prime, left parenthesis, x, right parenthesis, divided by, v, prime, left parenthesis, x, right parenthesis, end fraction exists, then the two limits are equal:

Let's find, for example, $\displaystyle\lim_{x\to 0}\dfrac{7x-\sin(x)}{x^2+\sin(3x)}$limit, start subscript, x, \to, 0, end subscript, start fraction, 7, x, minus, sine, left parenthesis, x, right parenthesis, divided by, x, squared, plus, sine, left parenthesis, 3, x, right parenthesis, end fraction.

Substituting $x=0$x, equals, 0 into $\dfrac{7x-\sin(x)}{x^2+\sin(3x)}$start fraction, 7, x, minus, sine, left parenthesis, x, right parenthesis, divided by, x, squared, plus, sine, left parenthesis, 3, x, right parenthesis, end fraction results in the indeterminate form $\dfrac{0}{0}$start fraction, 0, divided by, 0, end fraction. So let's use L’Hôpital’s rule.

Note that we were only able to use L’Hôpital’s rule because the limit $\displaystyle\lim_{x\to 0}\dfrac{\dfrac{d}{dx}[7x-\sin(x)]}{\dfrac{d}{dx}[x^2+\sin(3x)]}$limit, start subscript, x, \to, 0, end subscript, start fraction, start fraction, d, divided by, d, x, end fraction, open bracket, 7, x, minus, sine, left parenthesis, x, right parenthesis, close bracket, divided by, start fraction, d, divided by, d, x, end fraction, open bracket, x, squared, plus, sine, left parenthesis, 3, x, right parenthesis, close bracket, end fraction actually exists.

Let's find, for example, $\displaystyle\lim_{x\to 0}(1+2x)^{^{\LARGE\frac{1}{\sin(x)}}}$limit, start subscript, x, \to, 0, end subscript, left parenthesis, 1, plus, 2, x, right parenthesis, start superscript, start superscript, start fraction, 1, divided by, sine, left parenthesis, x, right parenthesis, end fraction, end superscript, end superscript. Substituting $x=0$x, equals, 0 into the expression results in the indeterminate form $1^{^{\Large\infty}}$1, start superscript, start superscript, infinity, end superscript, end superscript.

To make the expression easier to analyze, let's take its natural log (this is a common trick when dealing with composite exponential functions). In other words, letting $y=(1+2x)^{^{\LARGE\frac{1}{\sin(x)}}}$y, equals, left parenthesis, 1, plus, 2, x, right parenthesis, start superscript, start superscript, start fraction, 1, divided by, sine, left parenthesis, x, right parenthesis, end fraction, end superscript, end superscript, we will find $\displaystyle\lim_{x\to 0}\ln(y)$limit, start subscript, x, \to, 0, end subscript, natural log, left parenthesis, y, right parenthesis. Once we find it, we will be able to find $\displaystyle\lim_{x\to 0}y$limit, start subscript, x, \to, 0, end subscript, y.

$\ln(y) =\dfrac{\ln(1+2x)}{\sin(x)}$natural log, left parenthesis, y, right parenthesis, equals, start fraction, natural log, left parenthesis, 1, plus, 2, x, right parenthesis, divided by, sine, left parenthesis, x, right parenthesis, end fraction

Substituting $x=0$x, equals, 0 into $\dfrac{\ln(1+2x)}{\sin(x)}$start fraction, natural log, left parenthesis, 1, plus, 2, x, right parenthesis, divided by, sine, left parenthesis, x, right parenthesis, end fraction results in the indeterminate form $\dfrac{0}{0}$start fraction, 0, divided by, 0, end fraction, so now it's L’Hôpital’s rule's turn to help us with our quest!

We found that $\displaystyle\lim_{x\to 0}\ln(y)=2$limit, start subscript, x, \to, 0, end subscript, natural log, left parenthesis, y, right parenthesis, equals, 2, which means $\displaystyle\lim_{x\to 0}y=e^2$limit, start subscript, x, \to, 0, end subscript, y, equals, e, squared.