lim _(substack(x arrow 0 0)) (5^x-3^x-2^x+1)/(x^2)

To find the limit of the given expression as \( x \) approaches 0, we can use the Taylor series expansion for each term in the numerator and the denominator.<br /><br />The Taylor series expansion for \( 5^x \) around \( x = 0 \) is:<br />\[ 5^x = 1 + x \ln(5) + \frac{x^2}{2} \ln(5) + \frac{x^3}{6} \ln(5) + \ldots \]<br /><br />The Taylor series expansion for \( 3^x \) around \( x = 0 \) is:<br />\[ 3^x = 1 + x \ln(3) + \frac{x^2}{2} \ln(3) + \frac{x^3}{6} \ln(3) + \ldots \]<br /><br />The Taylor series expansion for \( 2^x \) around \( x = 0 \) is:<br />\[ 2^x = 1 + x \ln(2) + \frac{x^2}{2} \ln(2) + \frac{x^3}{6} \ln(2) + \ldots \]<br /><br />The Taylor series expansion for \( x^2 \) around \( x = 0 \) is:<br />\[ x^2 = x^2 \]<br /><br />Now, let's substitute these expansions into the given expression:<br /><br />\[ \lim _{\substack{x \rightarrow 0 \\ 0}} \frac{5^{x}-3^{x}-2^{x}+1}{x^{2}} = \lim _{\substack{x \rightarrow 0 \\ 0}} \frac{(1 + x \ln(5) + \frac{x^2}{2} \ln(5) + \ldots) - (1 + x \ln(3) + \frac{x^2}{2} \ln(3) + \ldots) - (1 + x \ln(2) + \frac{x^2}{2} \ln(2) + \ldots) + 1}{x^2} \]<br /><br />Simplifying the numerator, we get:<br /><br />\[ \lim _{\substack{x \rightarrow 0 \\ 0}} \frac{x \ln(5) - x \ln(3) - x \ln(2)}{x^2} \]<br /><br />Now, we can cancel out the common factor of \( x \) in the numerator and denominator:<br /><br />\[ \lim _{\substack{x \rightarrow 0 \\ 0}} \frac{\ln(5) - \ln(3) - \ln(2)}{x} \]<br /><br />Finally, substituting \( x = 0 \) into the expression, we get:<br /><br />\[ \lim _{\substack{x \rightarrow 0 \\ 0}} \frac{\ln(5) - \ln(3) - \ln(2)}{x} = \frac{\ln(5) - \ln(3) - \ln(2)}{0} \]<br /><br />Since the denominator is zero, the limit is undefined. Therefore, the correct answer is that the limit does not exist.