Problem 303
Question
$$ \lim _{x \rightarrow 0} \frac{a^{x}-1-x \ln a}{x^{2}}\left\\{\text { Ans. } \frac{\ln ^{2} a}{2}\right\\} $$
Step-by-Step Solution
Verified Answer
The short answer to the question is:
\(\lim_{x \rightarrow 0} \frac{a^{x}-1-x \ln a}{x^{2}} = \boxed{\frac{\ln^{2}a}{2}}\)
1Step 1: Calculate the derivatives of the given function
1. Calculate the derivative of \(a^x\) with respect to x: \(\frac{d}{dx}(a^x) = a^x \ln a\)
2. Calculate the derivative of \(-1\) with respect to x: \(\frac{d}{dx}(-1) = 0\)
3. Calculate the derivative of \(-x\ln a\) with respect to x: \(\frac{d}{dx}(-x\ln a) = -\ln a\)
2Step 2: Apply L'Hopital's rule
Applying L'Hopital's rule, we need to find the limit of the new function formed by the derivatives of the numerator and denominator as x approaches 0:
\(\lim_{x \to 0} \frac{a^x \ln a - 0 - (-\ln a)}{2x}\)
3Step 3: Simplify the new function
Now, let's simplify the new function:
\(\lim_{x \to 0} \frac{a^x\ln a + \ln a}{2x}\)
Factor out the \(\ln a\) from the numerator:
\(\lim_{x \to 0} \frac{\ln a(a^x + 1)}{2x}\)
4Step 4: Evaluate the limit
As x approaches 0, \(a^x\) approaches 1 (because anything raised to the power of 0 equals 1). So, the function becomes:
\(\lim_{x \to 0} \frac{\ln a(1 + 1)}{2x}\)
Since there's no longer any indeterminate form:
\(\frac{\ln a(2)}{2(0)} = \frac{2\ln a}{0}\)
However, division by zero is undefined. To avoid this issue, let's rewrite the function and apply L'Hopital's rule once more:
\(\lim_{x \to 0} \frac{\ln a(a^x + 1)}{2x}\)
After applying L'Hopital's rule again, we get:
\(\lim_{x \to 0} \frac{a^x\ln^2 a}{2}\)
Now, as x approaches 0, \(a^x\) approaches 1:
\(\frac{1\cdot\ln^2 a}{2} = \frac{\ln^2 a}{2}\)
So, the answer is:
\(\lim_{x \rightarrow 0} \frac{a^{x}-1-x \ln a}{x^{2}} = \frac{\ln^{2}a}{2}\)
Key Concepts
Limit of a functionDerivativeExponential functionLogarithmic function
Limit of a function
Understanding the limit of a function is about finding the value that a function approaches as the input approaches a certain point. In this problem, we’re dealing with the limit as \(x\) approaches 0.
Limits help us understand behaviors at points where functions aren't directly defined or are difficult to evaluate. They are foundational in calculus, especially when we deal with indeterminate forms like \(\frac{0}{0}\).
**Key Points:**
Limits help us understand behaviors at points where functions aren't directly defined or are difficult to evaluate. They are foundational in calculus, especially when we deal with indeterminate forms like \(\frac{0}{0}\).
**Key Points:**
- Limits can reveal the behavior of functions near certain points.
- They are essential in defining derivatives and integrals.
- Used to handle indeterminate forms through techniques like L'Hopital's Rule.
Derivative
The derivative of a function measures how a function changes as its input changes. It tells us the rate of change or the slope of the function at any point.
In this problem, derivatives simplify the expressions, making it easier to apply L'Hopital's Rule. By focusing on differential calculus, we determine how slight changes in \(x\) affect the overall expression.
**Key Points:**
In this problem, derivatives simplify the expressions, making it easier to apply L'Hopital's Rule. By focusing on differential calculus, we determine how slight changes in \(x\) affect the overall expression.
**Key Points:**
- The derivative of \(a^x\) is \(a^x \ln a\).
- Constants have derivatives of zero; hence the derivative of \(-1\) is 0.
- The derivative of \(-x\ln a\) is \(-\ln a\).
Exponential function
Exponential functions, like \(a^x\), are crucial because they describe growth and decay processes. In this scenario, \(a^x\) changes as \(x\) changes, and the function's behavior near zero is essential for understanding the overall limit.
**Key Properties:**
**Key Properties:**
- Base \(a\) must be greater than 0.
- As \(x\) approaches 0, \(a^x\) approaches 1 for any positive \(a\).
- The rate of change of an exponential function is proportional to its value, shown in its derivative \(a^x \ln a\).
Logarithmic function
Logarithmic functions, often written as \(\ln a\), are inverses of exponential functions. They unravel exponential changes, showing how fast growth or decay occurs.
In this context, \(\ln a\) helps simplify expressions involving exponential functions and plays a significant role in calculating limits and derivatives.
**Key Properties:**
In this context, \(\ln a\) helps simplify expressions involving exponential functions and plays a significant role in calculating limits and derivatives.
**Key Properties:**
- \(\ln a\) corresponds to the exponent needed for \(e\) (natural base) to equal \(a\).
- They transform products into sums, aiding in simplifying expressions.
- The derivative of \(\ln x\) is \(\frac{1}{x}\), illustrating how logarithmic functions slow as \(x\) increases.
Other exercises in this chapter
Problem 301
$$ \lim _{x \rightarrow 0}(\cos x+a \sin b x)^{\frac{1}{x}}\left\\{\text { Ans. } e^{a b}\right\\} $$
View solution Problem 302
$$ \lim _{x \rightarrow \infty} \cos \sqrt{x+1}-\cos \sqrt{x} \quad\\{\text { Ans. } 0\\} $$
View solution Problem 304
$$ \lim _{x \rightarrow \infty} \frac{3^{x}+4^{x}}{5^{x}+6^{x}}\\{\text { Ans. } 0\\} $$
View solution Problem 305
$$ \lim _{x \rightarrow 0} \frac{\cos ^{-1}(1-x)}{\sqrt{x}}\\{\text { Ans. } \sqrt{2}\\} $$
View solution