Problem 357
Question
$$ \lim _{x \rightarrow 0} \frac{a^{x}-b^{x}}{x \sqrt{1-x^{2}}}\left\\{\text { Ans. } \ln \frac{a}{b}\right\\} $$
Step-by-Step Solution
Verified Answer
The short answer for the given limit problem is \(\lim_{x \rightarrow 0} \frac{a^{x}-b^{x}}{x \sqrt{1-x^{2}}} = \ln{\frac{a}{b}}\).
1Step 1: Identify the indeterminate form to apply L'Hospital's rule
We first need to check if the limit has an indeterminate form (i.e., \(\frac{0}{0}\)) when x approaches 0, so we can apply L'Hospital's rule to find the limit.
When \(x = 0\), we have:
Numerator: \(a^0 - b^0 = 1-1 = 0\)
Denominator: \(0\sqrt{1-0^2} = 0\)
So, we have an indeterminate form of \(\frac{0}{0}\), which allows us to apply L'Hospital's rule.
2Step 2: Apply L'Hospital's rule
To apply L'Hospital's rule, we need to find the derivatives of the numerator and the denominator with respect to \(x\). Let's find those derivatives:
Numerator derivative: \(\frac{d}{dx} (a^x - b^x) = a^x\ln{a} - b^x\ln{b}\)
Denominator derivative: \(\frac{d}{dx} (x\sqrt{1-x^2}) = \sqrt{1-x^2} + x\frac{-2x}{2\sqrt{1-x^2}} = \sqrt{1-x^2} - x^2\sqrt{1-x^2}\)
Now, we will apply L'Hospital's rule:
\(\lim_{x\to 0} \frac{a^x\ln{a} - b^x\ln{b}}{\sqrt{1-x^2} - x^2\sqrt{1-x^2}}\)
3Step 3: Evaluate the limit
Let's substitute \(x\) with 0 in the new function after applying L'Hospital's rule:
\(\lim_{x\to 0} \frac{a^0\ln{a} - b^0\ln{b}}{\sqrt{1-0^2} - 0^2\sqrt{1-0^2}} = \frac{\ln{a} - \ln{b}}{1}\)
Now, we can use the property of logarithms to combine the logs:
\(\frac{\ln{a} - \ln{b}}{1} = \frac{\ln{a} - \ln{b}}{1} = \ln{\frac{a}{b}}\)
Thus, the required limit is:
\(\lim_{x \rightarrow 0} \frac{a^{x}-b^{x}}{x \sqrt{1-x^{2}}} = \ln{\frac{a}{b}}\)
Key Concepts
Indeterminate formsLimitsDifferential calculus
Indeterminate forms
In calculus, indeterminate forms arise when we evaluate limits that don't lead to a straightforward answer, like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These forms tell us that we need to dig deeper, to find the actual limit by alternative methods.
In the given problem, as \( x \) approaches 0, both the numerator \( a^x - b^x \) and the denominator \( x\sqrt{1-x^2} \) evaluate to 0, leading to the indeterminate form \( \frac{0}{0} \). This signals us to use a rule of calculus called L'Hospital's Rule.
Key points about indeterminate forms:
In the given problem, as \( x \) approaches 0, both the numerator \( a^x - b^x \) and the denominator \( x\sqrt{1-x^2} \) evaluate to 0, leading to the indeterminate form \( \frac{0}{0} \). This signals us to use a rule of calculus called L'Hospital's Rule.
Key points about indeterminate forms:
- They indicate that straightforward substitution in limits won’t work.
- L'Hospital's Rule is often used to solve these forms.
- They require careful attention to both the numerator and denominator.
Limits
A limit examines the behavior of a function as the input approaches a certain value. Understanding limits is fundamental in calculus, as they form the basis of derivatives, integrals, and continuity.
In our problem, we need to find the limit of \( \frac{a^x - b^x}{x\sqrt{1-x^2}} \) as \( x \) approaches 0. Since both the numerator and the denominator approach 0, we apply tools like differentiation to explore under what conditions the limit exists.
Important points about limits:
In our problem, we need to find the limit of \( \frac{a^x - b^x}{x\sqrt{1-x^2}} \) as \( x \) approaches 0. Since both the numerator and the denominator approach 0, we apply tools like differentiation to explore under what conditions the limit exists.
Important points about limits:
- They can describe forces such as speed or rate changes over time.
- Calculating a limit involves substituting and simplifying expressions to find outputs close to the target point.
- They can yield finite numbers, infinity, or fail to exist.
Differential calculus
Differential calculus focuses on the study of derivatives, which represent the rate of change of a function. It is a key tool in finding exact values for indeterminate forms.
L'Hospital's Rule leverages differential calculus by stating that if a limit results in \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), we can evaluate the limit of the derivatives of the numerator and the denominator. In our example, we first compute the derivatives:
L'Hospital's Rule leverages differential calculus by stating that if a limit results in \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), we can evaluate the limit of the derivatives of the numerator and the denominator. In our example, we first compute the derivatives:
- The derivative of \( a^x - b^x \) with respect to \( x \) is \( a^x \ln a - b^x \ln b \).
- The derivative of \( x \sqrt{1-x^2} \) is derived using the product and chain rules, resulting in \( \sqrt{1-x^2} - x^2\sqrt{1-x^2} \).
Other exercises in this chapter
Problem 355
$$ \lim _{x \rightarrow \infty} \frac{\pi-2 \tan ^{-1} x}{\ln \left(1+\frac{1}{x}\right)}\\{\text { Ans. } 2\\} $$
View solution Problem 356
$$ \lim _{x \rightarrow 0} \frac{a^{x}-b^{x}}{c^{x}-d^{x}}\left\\{\text { Ans. } \frac{\ln \frac{a}{b}}{\ln \frac{c}{d}}\right\\} $$
View solution Problem 358
$$ \lim _{x \rightarrow a} \frac{\cos x \ln (x-a)}{\ln \left(e^{x}-e^{a}\right)}\\{\text { Ans. } \cos a\\} $$
View solution Problem 359
$$ \lim _{x \rightarrow 0} \frac{e^{\tan x}-e^{x}}{\tan x-x}\\{\text { Ans. } 1\\} $$
View solution