Indeterminate form limits question

Question

$$\lim_{x\to 0}\frac{10^x - 2^x - 5 ^ x + 1 } {x\tan x} $$ This is an indeterminate limit. I want help in solving this problem. Thanks in advance

Answer 1

If we replace in the expression $-1$ by $1$ we have in this case the indeterminate form $\frac 00$ and we calculate the limit using the Taylor series:

$$\frac{10^x - 2^x - 5 ^ x + 1 } {x\tan x}=\frac{\exp(x\log10) -\exp(x\log2) - \exp(x\log5) + 1 } {x\tan x}\sim_0\frac{1+x\log10+\frac{x^2\log^210}{2}-1-x\log2-\frac{x^2\log^22}{2}-1-x\log5-\frac{x^2\log^25}{2}+1}{x^2}\sim_0\frac{\log^210-\log^22-\log^25}{2}$$

Answer 2

$$\frac{10^x - 2^x - 5 ^ x + 1 } {x\tan x}$$

$$=\frac{(5^x-1)(2^x-1)}{x\tan x}$$

$$=\frac{\dfrac{5^x-1}x\cdot\dfrac{2^x-1}x\cdot\cos x}{\dfrac{\sin x}x}$$

Now as $\displaystyle a^h=\left(e^{\ln a}\right)^h=e^{h\cdot\ln a}$,

$\displaystyle\lim_{h\to0}\frac{a^h-1}h=\ln a\cdot\lim_{h\to0}\frac{e^{h\ln a}-1}{h\ln a}=\ln a$