Please answer the question providing all steps and clear writing so I can understand. Thanks.You can see as a Reference book "James Stewart Calculus 8th Edition" to see similar solutions. To solve the limit problem, we need to evaluate the following expression:\[ \lim_{{x \to 0}} \frac{5^x - 1}{2x} \]Here’s a step-by-step explanation on how to approach this problem:1. **Direct Substitution**: Initially, you might think to substitute \( x = 0 \) directly into the expression. However, doing so results in the indeterminate form \(\frac{0}{0}\). Therefore, direct substitution doesn't work in this case and alternative methods should be considered.2. **Applying L'Hôpital's Rule**: When faced with an indeterminate form such as \(\frac{0}{0}\), L'Hôpital's Rule can be used. L'Hôpital's Rule states that if \(\lim_{{x \to c}} f(x)/g(x) = 0/0\) or \(\pm \infty/\pm \infty\), then: \[ \lim_{{x \to c}} \frac{f(x)}{g(x)} = \lim_{{x \to c}} \frac{f'(x)}{g'(x)} \] provided that the limit on the right-hand side exists.3. **Differentiate the Numerator and Denominator**: The numerator \( 5^x - 1 \) and the denominator \( 2x \) need to be differentiated separately: - The derivative of \( 5^x \) is: \( 5^x \ln 5 \) - The derivative of \( -1 \) is: \( 0 \) - The derivative of \( 2x \) is: \( 2 \)4. **Substitute the Derivatives**: Applying L'Hôpital's Rule: \[ \lim_{{x \to 0}} \frac{5^x - 1}{2x} = \lim_{{x \to 0}} \frac{5^x \ln 5}{2} \]5. **Evaluate the Limit**: Now, substituting \( x = 0 \) into \(\frac{5^x \ln 5}{2}\) gives: \[ \frac{5^0 \ln 5}{2} = \frac{1 \cd

Name: Please answer the question providing all steps and clear writing so I
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Description: Please answer the question providing all steps and clear writing so I can understand. Thanks.You can see as a Reference book "James Stewart Calculus 8th Edition" to see similar solutions. To solve the limit problem, we need to evaluate the following