Use direct substitution method to find the limit of the function as $x$ approaches zero in this calculus problem.

$\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{5^x+5^{-x}-2}{x^2}}$

$= \,\,\,$ $\dfrac{5^0+5^{-0}-2}{{(0)}^2}$

$= \,\,\,$ $\dfrac{5^0+5^{0}-2}{0}$

$= \,\,\,$ $\dfrac{1+1-2}{0}$

$= \,\,\,$ $\dfrac{2-2}{0}$

$= \,\,\,$ $\dfrac{0}{0}$

In this limit problem, it is evaluated that the limit of the function is indeterminate. So, it is not possible to evaluate the limit of the given function by the direct substitution method. Now, let’s think to evaluate the limit of algebraic function in another mathematical approach.

$\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{5^x+5^{-x}-2}{x^2}}$

The expression in the denominator is in square form and the algebraic expression in the numerator should also be in the same form for balancing the denominator. Therefore, let us try to convert the exponential algebraic expression in square form. Mathematically, it is possible in this case.

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{5^x+\dfrac{1}{5^x}-2}{x^2}}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{5^x+\dfrac{1}{5^x}-2}{x^2}}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\dfrac{5^x \times 5^x+1-2 \times 5^x}{5^x}}{x^2}}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{{(5^{x})}^2+1-2 \times 5^x}{5^x \times x^2}}$

If the number $1$ is written as square of $1$ in the numerator, then the algebraic expression in the numerator represents the expansion of the square of difference of two terms identity.

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{{(5^{x})}^2+1^2-2 \times 5^x \times 1}{5^x \times x^2}}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{{(5^x-1)}^2}{5^x \times x^2}}$

The algebraic function has to factorise as two functions and it is useful to find limit of the function in this problem.

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg[\dfrac{1}{5^x} \times \dfrac{{(5^x-1)}^2}{x^2}\Bigg]}$

Now, use product rule of limits to find the limit of product of functions by the product of their limits.

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1}{5^x}}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{{(5^x-1)}^2}{x^2}}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1}{5^x}}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{{(5^x-1)}^2}{x^2}}$

Now, find the limit of the first function by using direct substitution method.

$= \,\,\,$ $\dfrac{1}{5^0}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{{(5^x-1)}^2}{x^2}}$

$= \,\,\,$ $\dfrac{1}{1}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{{(5^x-1)}^2}{x^2}}$

$= \,\,\,$ $1$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{{(5^x-1)}^2}{x^2}}$

Now, concentrate on finding the limit of the following algebraic function.

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{{(5^x-1)}^2}{x^2}}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize {\Bigg(\dfrac{5^x-1}{x}\Bigg)}^2}$

This mathematical function represents the limit of an exponential function but its exponent is a constant. So, it can be further simplified as per constant power rule of limits.

$= \,\,\,$ $\Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize {\dfrac{5^x-1}{x}\Bigg)}^2}$

According to the limit of (a^{x}-1)/x as x approaches 0 rule, the limit of exponential function can be evaluated.

$= \,\,\,$ ${(\log_{e}{(5)})}^2$

$= \,\,\,$ ${(\ln{(5)})}^2$

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