Evaluating the Limit of an Exponential Function Expression

The problem asks us to find the limit of the function \(\rm \displaystyle\lim_{x \rightarrow 0} \dfrac{3^x + 3^{-x}-2}{x}\).

First, let's evaluate the function at \(x = 0\) to see if it's an indeterminate form.

Numerator at \(x=0\): \(3^0 + 3^{-0} - 2 = 1 + 1 - 2 = 0\)

Denominator at \(x=0\): \(0\)

Since we get the form \(\dfrac{0}{0}\) as \(x \rightarrow 0\), this is an indeterminate form. We can use methods like L'Hopital's Rule or standard limit formulas to evaluate this limit.

Method 1: Using L'Hopital's Rule

L'Hopital's Rule states that if \(\displaystyle\lim_{x \rightarrow c} \dfrac{f(x)}{g(x)}\) is of the form \(\dfrac{0}{0}\) or \(\dfrac{\infty}{\infty}\), then \(\displaystyle\lim_{x \rightarrow c} \dfrac{f(x)}{g(x)} = \lim_{x \rightarrow c} \dfrac{f'(x)}{g'(x)}\), provided the latter limit exists.

In our case, \(f(x) = 3^x + 3^{-x} - 2\) and \(g(x) = x\).

We need to find the derivatives of \(f(x)\) and \(g(x)\).

• The derivative of \(3^x\) is \(3^x \ln 3\).
• The derivative of \(3^{-x}\) is \(3^{-x} (\ln 3) \cdot (-1) = -3^{-x} \ln 3\) (using the chain rule).
• The derivative of \(-2\) is \(0\).
• The derivative of \(x\) is \(1\).

So, \(f'(x) = \dfrac{d}{dx}(3^x + 3^{-x} - 2) = 3^x \ln 3 - 3^{-x} \ln 3 = \ln 3 (3^x - 3^{-x})\).

And \(g'(x) = \dfrac{d}{dx}(x) = 1\).

Now, we apply L'Hopital's Rule:

\(\displaystyle\lim_{x \rightarrow 0} \dfrac{3^x + 3^{-x}-2}{x} = \lim_{x \rightarrow 0} \dfrac{\ln 3 (3^x - 3^{-x})}{1}\)

Substitute \(x=0\) into the new expression:

\(\ln 3 (3^0 - 3^{-0}) = \ln 3 (1 - 1) = \ln 3 \cdot 0 = 0\).

Method 2: Using Standard Limit Formulas

Recall the standard limit formula: \(\displaystyle\lim_{x \rightarrow 0} \dfrac{a^x - 1}{x} = \ln a\).

We can rewrite the numerator of the given expression: \(3^x + 3^{-x} - 2 = (3^x - 1) + (3^{-x} - 1)\).

So, the limit becomes:

\(\displaystyle\lim_{x \rightarrow 0} \dfrac{(3^x - 1) + (3^{-x} - 1)}{x} = \lim_{x \rightarrow 0} \left( \dfrac{3^x - 1}{x} + \dfrac{3^{-x} - 1}{x} \right)\)

We can evaluate the two parts separately:

Part 1: \(\displaystyle\lim_{x \rightarrow 0} \dfrac{3^x - 1}{x}\)

Using the standard formula with \(a=3\), this limit is \(\ln 3\).

Part 2: \(\displaystyle\lim_{x \rightarrow 0} \dfrac{3^{-x} - 1}{x}\)

Let \(y = -x\). As \(x \rightarrow 0\), \(y \rightarrow 0\). Also, \(x = -y\).

Substitute these into the limit:

\(\displaystyle\lim_{y \rightarrow 0} \dfrac{3^y - 1}{-y} = -\lim_{y \rightarrow 0} \dfrac{3^y - 1}{y}\)

Using the standard formula with \(a=3\), \(\displaystyle\lim_{y \rightarrow 0} \dfrac{3^y - 1}{y} = \ln 3\).

So, Part 2 limit is \(-\ln 3\).

Combining the results for Part 1 and Part 2:

The total limit is \(\ln 3 + (-\ln 3) = \ln 3 - \ln 3 = 0\).

Conclusion

Both methods show that the limit of \(\rm \displaystyle\lim_{x \rightarrow 0} \dfrac{3^x + 3^{-x}-2}{x}\) is 0.

Revision Table: Key Concepts for Limits

Additional Information: Exponential Function Derivatives and Limits

The derivative of an exponential function \(f(x) = a^x\), where \(a > 0\) and \(a \neq 1\), is \(f'(x) = a^x \ln a\). This is a fundamental result in differential calculus.

The standard limit \(\displaystyle\lim_{x \rightarrow 0} \dfrac{a^x - 1}{x} = \ln a\) can be derived using the definition of the derivative. Consider the derivative of \(f(x) = a^x\) at \(x=0\).

\(f'(0) = \displaystyle\lim_{h \rightarrow 0} \dfrac{f(0+h) - f(0)}{h} = \lim_{h \rightarrow 0} \dfrac{a^{0+h} - a^0}{h} = \lim_{h \rightarrow 0} \dfrac{a^h - 1}{h}\).

Since \(f'(x) = a^x \ln a\), we have \(f'(0) = a^0 \ln a = 1 \cdot \ln a = \ln a\).

Therefore, \(\displaystyle\lim_{h \rightarrow 0} \dfrac{a^h - 1}{h} = \ln a\), which is the standard limit formula we used. Replacing the variable \(h\) with \(x\) gives the form \(\displaystyle\lim_{x \rightarrow 0} \dfrac{a^x - 1}{x} = \ln a\).