# Math Help - L'Hospital's rule

1. ## L'Hospital's rule

Can anyone explain the procedures in solving these two problems using L'Hospital's rule?

1. lim x-> 0+ (tan 2x)^x

2. lim x-> 1 (2 - x)^tan((pi)x/2)

2. If the function is of indeterminate form, then take the derivative and find the limit of that. Read up, a tad on L'Hop's rule, it is not a hard rule you just need to know when to apply it

3. Hello, Eternal!

Here's the second one . . .

$(2)\;\;\lim_{x\to1}\,(2 - x)^{\tan(\frac{\pi}{2}x)}$

Let: . $y \:=\:(2-x)^{\tan(\frac{\pi}{2}x)}$

Take logs: . $\ln(y) \;=\;\ln\left[(2-x)^{\tan(\frac{\pi}{2}x)}\right] \;=\;\tan(\tfrac{\pi}{2}x)\cdot\ln(2-x) \;=\;\frac{\ln(2-x)}{\cot(\frac{\pi}{2}x)} \quad\rightarrow\quad \frac{0}{0}$

Apply L'Hopital: . $\ln(y) \;=\;\frac{\frac{-1}{2-x}} {-\frac{\pi}{2}\csc^2(\frac{\pi}{2}x)} \;=\;\frac{2\sin^2(\frac{\pi}{2}x)}{\pi(2-x)}$

Take limits: . $\lim_{x\to1}\,\ln(y) \;=\;\lim_{x\to1}\,\frac{2\sin^2(\frac{\pi}{2}x)}{ \pi(2-x)} \;=\;\frac{2\cdot1^2}{\pi(1)} \;=\;\frac{2}{\pi}$

We have: . $\ln(y) \:=\:\frac{2}{\pi}$

Therefore: . $y \;=\;e^{\frac{2}{\pi}}$

4. Hello again, Eternal!

It took a while to get the first one.
I hope I'm right . . .

$(1)\;\;\lim_{x\to0^+}(\tan 2x)^x$
As given, the limit goes to $0^0$ . . . an indeterminate form.

Let: . $y \:=\:(\tan2x)^x$

Take logs: . $\ln(y) \:=\:\ln(\tan2x)^x \:=\;x\cdot\ln(\tan2x) \;=\;\frac{\ln(\tan2x)}{\frac{1}{x}} \;=\;\frac{\ln(\tan2x)}{x^{-1}} \quad \to \quad \frac{\text{-}\infty}{\infty}$

Apply L'Hopital: . $\frac{\;\dfrac{2\sec^2\!2x}{\tan2x}\;}{-x^{-2}} \;=\;\frac{-2x^2\sec^2\!2x}{\tan2x} \;=\; \frac{-2x^2}{\sin2x\cos2x} \;=\;\frac{-4x^2}{2\sin2x\cos2x} \;=\;\frac{-4x^2}{\sin4x} \quad\to \quad \frac{0}{0}$

Apply L'Hopital: . $\frac{-8x}{4\cos4x} \;=\;\frac{-2x}{\cos4x}$

Take the limit: . $\lim_{x\to0^+}\frac{-2x}{\cos4x} \;=\;\frac{0}{1} \;=\;0$

We have: . $\lim_{x\to0^+}\bigg[\ln(y)\bigg] \:=\:0$,

. . therefore: . $\lim_{x\to0^+}\,y \;=\;e^0 \;=\;1$