How do I solve this limit without L'Hospital's rule?

Question

I want to evaluate this limit: $$\lim_{x\to1}\frac{\log_4(x+1)-0.5}{x-1}$$

I normally solve similar limits by somehow getting $\ln$ and $(1+x)^{1/x}$ to cancel out, but I don't know what to do here. I can get it to the form: $$\lim_{x\to0}\ln(x+2)^{1/x}-\frac{0.5}{x}\ln4$$

but it also isn't of any use because of the $2$ in the left $\ln$ and because of the $x$ that is still left on the right. So what can I do to solve this? I don't want to use L'Hospital, derivatives or any series expansions.

Answer 1

$$\lim_{x \to 1} \frac{\log_{4}{x+1} - 0,5}{x-1}= \lim_{x \to 1} \frac{\log_{4}{x+1} - \log_{4}{2}}{x-1}= \lim_{x \to 1} \frac{\log_{4}{\frac{x+1}{2}}}{x-1}= \lim_{x \to 0} \log_{4}{(1+\frac{x}{2})^{\frac{1}{x}}}= \frac{1}{2}\log_{4}e = \frac{1}{2 \ln{4}}$$

Answer 2

The limit is a derivative. Consider the function $f(t) = \log_4(t + 1)$. Then $f'(1)$ is, by definition of derivative, equal to $$ \lim_{x\to 1}\frac{f(x) - f(1)}{x-1} = \lim_{x\to 1}\frac{\log_4(x+1) - 0.5}{x-1} $$ Using chain rules and other derivative tools, we can find this derivative in other ways. As $\log_4(t+1) = \frac{\ln(t+1)}{\ln 4}$, we get $$ f'(t) = \frac{1}{\ln 4}\frac{1}{t+1} $$ Now just insert $t = 1$ into this, and you're done.

Answer 3

$$\lim_{x\to 1}\frac{\log_4(x+1) - 0.5}{x-1}=\lim_{x\to 0}\frac{\log_4(x+2) - 0.5}x=\frac12\lim_{x\to 0}\frac{\log_4(\frac x2+1)}{\frac x2} \\=\frac1{2\log 4}\lim_{x\to 0}\frac{\log(x+1)}x.$$

Now you have a known limit.