I was given the following problem and told to use L'Hospital's rule: $$\lim_{x\to2}(9-4x)^{\frac{1}{\tan(x-2)}}$$ But, I am confused - what makes this into indeterminate form?

• $\lim\limits_{x\to2}\tan(x-2)=0$ – J. W. Tanner Nov 24 '19 at 20:44
• It's of the indeterminate type $1^\infty$. The usual remedy is to look at its logarithm. Using the formula $\ln(a^b)=b\ln a$. It is often possible to calculate the limit of that with l'Hospital. – Jyrki Lahtonen Nov 24 '19 at 20:45
• @JyrkiLahtonen how is it $1^\infty$? – Burt Nov 24 '19 at 20:48
• As $x$ approaches $2$ from the right, $\tan(x-2)$ is approaching $0$ from the right, and as the denominator of the fraction $1/\tan(x-2)$ approaches zero from the right, the value of $1/\tan(x-2)$ approaches infinity. – Michael Burr Nov 24 '19 at 20:49
• In other words $$(9-4x)^{\frac1{\tan(x-2)}}=e^{\frac{\ln(9-4x)}{\tan(x-2)}}.$$ You've got $0/0$ in the exponent. Use L'Hospital on that. – Jyrki Lahtonen Nov 24 '19 at 20:55

Others have identified this as an $$1^\infty$$ indeterminate form. But perhaps it is unclear why this is an indeterminate form. Here are a few more limits having the same indeterminate form. \begin{align*} \lim_{x \rightarrow 0} (1+x)^{1/x} &= \mathrm{e} \text{.} \\ \lim_{x \rightarrow 0} (1+x^2)^{1/x} &= 1 \text{.} \\ \lim_{x \rightarrow 0} (1+x)^{1/x^2} &\text{ does not converge,} \\ \lim_{x \rightarrow 0^+} (1+x)^{1/x^2} &= \infty \text{, and} \\ \lim_{x \rightarrow 0^-} (1+x)^{1/x^2} &= 0 \text{.} \end{align*}
So if you are doing "simple analysis" to see if a limit has an easy to predict result, the limit of a "$$1^\infty$$" form depends on the details of how rapid the approach is to $$1$$ (and possibly from which side) and to $$\infty$$.
As @Jyrki mentions, the indeterminate form here is $$1^\infty$$ (which is one of the less common indeterminate forms to see in undergraduate calculus). The first step of solving such an indeterminate form is to take the logarithm and this shows a more common indeterminate form:
\begin{align*} \lim_{x\rightarrow 2}(9-4x)^{\frac{1}{\tan(x-2)}}&=e^{\ln\left(\lim_{x\rightarrow 2}(9-4x)^{\frac{1}{\tan(x-2)}}\right)}\\ &=e^{\lim_{x\rightarrow 2}\ln\left((9-4x)^{\frac{1}{\tan(x-2)}}\right)}\\ &=e^{\lim_{x\rightarrow 2}\frac{\ln(9-4x)}{\tan(x-2)}} \end{align*} As $$x$$ approaches $$2$$, the numerator approaches $$\ln(1)=0$$ and the denominator approaches $$\tan(0)=0$$. Therefore, we've transformed the limit into the (perhaps more common) indeterminate form $$\frac{0}{0}$$.