Is this procedure valid for finding the limit?

Question

$$\begin{align*} \lim_{x\to \pi/4} (1-\tan x )^{\cos x -\sin x } &= \lim_{x\to \pi/4} \frac{(1-\tan x )^{\cos x}}{(1-\tan x )^{\sin x }} \\ &= \lim_{x \to \pi/4} \frac{\ln (1-\tan x )^{\cos x } }{\ln (1-\tan x )^{\sin x }} \\ &= \lim_{x\to \pi/4} \frac{\cos x }{\sin x } \\ &= \lim_{x\to \pi/4} \cot x \\ &= \cot \frac{\pi}{4} \\ &= 1 \end{align*}$$

I didn't want to use L'Hopital so I tried this. I think I divided by 0. Help please (:

Answer 1

The taking of logarithms in the numerator and denominator has no logical basis. Doing such a thing is like saying $$\frac{2}{3} = \frac{\log 2}{\log 3},$$ which of course is absurd. Consequently, your calculation from that point forward is incorrect.

The other problem with the question is that the function $$f(x) = (1 - \tan x)^{\cos x - \sin x}$$ is not in general real-valued for $\pi/4 < x < \pi/2$, since in this interval, $1 - \tan x < 0$. Hence the two-sided limit, strictly speaking, does not exist. The one-sided limit (from below) does, however, which we show as follows:

Note that $$1 - \tan x = \frac{\cos x - \sin x}{\cos x}.$$ So $$f(x) = (g(x) \sec x)^{g(x)} = g(x)^{g(x)} (\sec x)^{g(x)},$$ where $g(x) = \cos x - \sin x$. As $x \to \pi/4^-$, $g(x) \to 0^+$ whereas $\sec x \to \sqrt{2}$, hence the limit of the second factor exists and equals $1$, allowing us to write $$\lim_{x \to \pi/4^-} f(x) = \lim_{x \to \pi/4^-} g(x)^{g(x)} \cdot \lim_{x \to \pi/4^-} (\sec x)^{g(x)} = \lim_{x \to \pi/4^-} g(x)^{g(x)}.$$ Then we merely conclude that the remaining limit can be regarded as a function of $g$ rather than $x$, i.e., $$\lim_{x \to \pi/4^-} g(x)^{g(x)} = \lim_{g \to 0^+} g^g = 1.$$

Answer 2

If you don't want to use L'Hopital you can use continuity.

\begin{equation} \lim_{x\to\pi/4}(1-\tan(x))^{\cos(x)-\sin(x)}=\exp\left(\lim_{x\to\pi/4}\ln\left((1-\tan(x))^{\cos(x)-\sin(x)}\right)\right)\\ =\exp\left(\lim_{x\to\pi/4}(\cos(x)-\sin(x))\cdot\ln\left((1-\tan(x))\right)\right) \end{equation}

Now, write $$\ln\left((1-\tan(x))\right)=\ln\left(\frac{\cos(x)-\sin(x)}{\cos(x)}\right)=\ln(\cos(x)-\sin(x))-\ln(\cos(x))$$

So we may calculate two distinct limits:

\begin{equation} \lim_{x\to\pi/4}(\cos(x)-\sin(x))\cdot\ln\left(\cos(x)-\sin(x)\right)=\lim_{y\to0}y\cdot\ln(y)=\lim_{y\to0}\ln\left(y^y\right)\\ \lim_{x\to\pi/4}(\cos(x)-\sin(x))\cdot\ln\left(\cos(x)\right)=0 \end{equation}

Since $y^y\to 1$ as $y \to 0$ (from the positives), the first limit is $0$, to the original limit is $\exp(0)=1$.

Answer 3

Whenever we have both base and exponent as variables it is best to take logs in order to evaluate the limit. Also note that the limit here must be as $x \to \pi/4^{-}$ so that $(1 - \tan x) > 0$. Thus if $L$ is the desired limit then we have \begin{align} \log L &= \log\left\{\lim_{x \to \pi/4^{-}}(1 - \tan x)^{\cos x - \sin x}\right\}\notag\\ &= \lim_{x \to \pi/4^{-}}\log(1 - \tan x)^{\cos x - \sin x}\text{ (via continuity of log)}\notag\\ &= \lim_{x \to \pi/4^{-}}(\cos x - \sin x)\log(1 - \tan x)\notag\\ &= \lim_{x \to \pi/4^{-}}(\cos x - \sin x)\{\log(\cos x - \sin x) - \log \cos x\}\notag\\ &= \lim_{x \to \pi/4^{-}}(\cos x - \sin x)\log(\cos x - \sin x) - (\cos x - \sin x)\log \cos x\notag\\ &= \lim_{x \to \pi/4^{-}}(\cos x - \sin x)\log(\cos x - \sin x) - 0\cdot \log 1\notag\\ &= \lim_{t \to 0^{+}}t\log t\text{ (putting }t = \cos x - \sin x)\notag\\ &= 0\notag \end{align} Hence $L = 1$. Here we have used the standard limit $$\lim_{t \to 0^{+}}t^{a}(\log t)^{b} = 0$$ for positive $a, b$.

Answer 4

HINT:

$$\lim_{x\to \pi/4} (1-\tan x )^{\cos x -\sin x }=\left[\lim_{x\to\pi/4}\left(1-\tan x\right)^{1-\tan x}\right]^{\lim_{x\to\pi/4}\cos x}$$

Set $1-\tan x=y$