# Solution verification:$\lim_{x\to 2}\frac{\ln(x-1)}{3^{x-2}-5^{-x+2}}$

Evaluate without L'Hospital:$$\lim_{x\to 2}\frac{\ln(x-1)}{3^{x-2}-5^{-x+2}}$$

My attempt:

I used: $$\lim_{f(x)\to 0}\frac{\ln(1+f(x))}{f(x)}=1\;\&\;\lim_{f(x)\to 0}\frac{a^{f(x)}-1}{f(x)}=\ln a$$

$$\begin{split} L &= \lim_{x\to 2} \frac{\ln(x-1)}{3^{x-2}-5^{-x+2}} \\ &= \lim_{x\to 2} \frac{\dfrac{\ln(1+(x-2))}{x-2}\cdot(x-2)} {(x-2)\cdot\dfrac{3^{x-2}-1+1-5^{-x+2}}{x-2}} \\ &= \lim_{x\to 2} \frac{\dfrac{\ln(1+(x-2))}{x-2}} {\dfrac{3^{x-2}-1}{x-2}+\dfrac{5^{2-x}-1}{2-x}} \\ &=\frac{1}{\ln3+\ln5} \\ &=\frac{1}{\ln(15)} \end{split}$$

Is this correct?

• looks good to me Jan 19, 2020 at 14:25
• @gt6989b, thank you for responding! Jan 19, 2020 at 14:26
• Thanks you for posting such a well-put and well-formatted question, very few people on the site do that :) If I could +1 again, I would Jan 19, 2020 at 14:29
• It is quite correct. You could have made it a bit shorter writing $\;3^{x-2}-5^{-x+2}=5^{-x+2}\bigl(15^{x-2}-1\bigr)$. Jan 19, 2020 at 14:30
• @gt6989b, thank you! I just enjoy this community and owe so much! I never thought the internet would be such a wonderful place for exchanging ideas. Jan 19, 2020 at 14:32

This is fine. Here is an alternative approach using taylor series:

First substitute $$x-2=y$$ to simplify it. Let the required limit be $$l$$. Then $$l = \lim_{y\to0}\left(\dfrac{\ln(1+y)}{3^y-5^{-y}}\right)$$ $$= \lim_{y\to0}\left(\dfrac{y-\dfrac{y^2}{2}+\cdots}{(1+y\ln3+\cdots)-(1-y\ln5+\cdots)}\right)$$ $$=\dfrac{1}{\ln3+\ln5}=\dfrac{1}{\ln15}$$

$$y=x-2;$$

$$\dfrac{\log (1+y)}{3^y-5^{-y}}$$;

Numerator :

$$f(y):=y\dfrac{\log (1+y)-\log 1}{y}$$

Denominator:

$$g(x)=\dfrac{15^y-1}{5^y}=$$

$$5^{-y}(15^y-1)=$$

$$5^{-y}(y\log 15)\dfrac{e^{y\log 15}-1}{y\log 15}$$;

$$\dfrac{f(x)}{g(x)}=$$

$$[\dfrac{\log (y+1)-\log 1}{y}]\cdot$$

$$[\dfrac{5^y}{\log 15}]$$

$$\big [\dfrac{1}{\dfrac{e^{y\log15}-1}{y\log 15}}\big ].$$

Take the limit $$y \rightarrow 0$$.