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Relevant wiki: Limits of functions - Problem solving

Given Limit will be $\displaystyle\lim_{x\rightarrow 0} \left(\dfrac{2^2\cdot\frac{\sin(x)}{x}\cdot2\cdot\frac{\sin^2(2x)}{2^2x^2}\left(\frac{(5^x-1)}{x}-\frac{(4^x-1}{x}\right)}{\frac{\tan(5x)}{(5x)}\cdot(5x)}\right)=\dfrac{8}{5}\ln\left(\dfrac{5}{4}\right)$

So $\mathfrak{q}=8,\mathfrak{r}=5,\mathfrak{t}=4$ their sum is $\boxed{17}$

I have used the standard limits like $\displaystyle\lim_{f(x)\rightarrow0} \frac{\sin(f(x))}{f(x)}=\displaystyle\lim_{f(x)\rightarrow0} \frac{\tan(f(x))}{f(x)}=1$ and $\displaystyle\lim_{f(x)\rightarrow0} \frac{a^{f(x)}-1}{f(x)} =\ln(a)$

For limits approaching zero and mutliplication we can replace sinx = x , tanx = x

Doing this we get

8/5 (5^x - 4^x / x)

Apply L'Hospitals rule to get answer