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Magazine Chapter 4: Problem 47
What happens if you try to use I'Hospital's Rule to find the limit? Evaluate the limit using another method. $$\lim _{x \rightarrow \infty} \frac{x}{\sqrt{x^{2}+1}}$$
Short Answer
Expert verified
Using L'Hospital's Rule is not helpful, but simplifying gives a limit of 1.
Step by step solution
01
Identify the Form
First, we need to check if using L'Hospital's Rule is applicable. The question involves \[ \lim _{x \rightarrow \infty} \frac{x}{\sqrt{x^{2}+1}} \]Finding the limit as \( x \) approaches infinity. The numerator is \( x \) and the denominator is \( \sqrt{x^2 + 1} \). Both tend towards infinity as \( x \rightarrow \infty \). Thus, we have an \( \frac{\infty}{\infty} \) indeterminate form.
02
Evaluate Using L'Hospital's Rule
Since the limit presents an \( \frac{\infty}{\infty} \) form, we could apply L'Hospital's Rule. This involves taking the derivative of the numerator and the derivative of the denominator:The derivative of the numerator \( x \) is \( 1 \). The derivative of the denominator \( \sqrt{x^2+1} \) is obtained using the chain rule and equals \( \frac{x}{\sqrt{x^2+1}} \).Applying L'Hospital's Rule, we get:\[\lim_{x\to\infty} \frac{1}{\frac{x}{\sqrt{x^2+1}}} = \lim_{x\to\infty} \frac{\sqrt{x^2+1}}{x}\]However, this takes us back to a similar \( \frac{\infty}{\infty} \) form, suggesting that L'Hospital's Rule is not simplifying the problem.
03
Simplify the Function
Instead of using L'Hospital's Rule, let's simplify the expression to find the limit. Begin by factoring \( x^2 \) out of the denominator:\[\lim _{x \rightarrow fty} \frac{x}{\sqrt{x^{2}+1}} = \lim _{x \rightarrow fty} \frac{x}{\sqrt{x^{2}(1+\frac{1}{x^{2}})}}\]This gives us:\[\lim _{x \rightarrow fty} \frac{x}{x\sqrt{1+\frac{1}{x^{2}}}}=\lim _{x \rightarrow fty} \frac{1}{\sqrt{1+\frac{1}{x^2}}}\]
04
Evaluate the Limit
As \( x \rightarrow fty \), the term \( \frac{1}{x^2} \rightarrow 0 \). Hence, \[ \sqrt{1+\frac{1}{x^2}} \rightarrow \sqrt{1} = 1\]Thus, the limit simplifies to:\[\lim_{x \rightarrow \infty} \frac{1}{\sqrt{1+\frac{1}{x^2}}} = 1\]
05
Conclusion: Final Answer
The limit of the function \( \lim _{x \rightarrow fty} \frac{x}{\sqrt{x^{2}+1}} \) is 1 after simplifying the function.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
L'Hospital's Rule
L'Hospital's Rule is a powerful tool in calculus used to simplify finding limits of indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). Suppose you have a limit of the form \( \lim_{x \to c} \frac{f(x)}{g(x)} \), where direct substitution results in an indeterminate form. In such cases, L'Hospital's Rule states that you can take the derivatives of the numerator and denominator:
- Find \( f'(x) \), the derivative of the numerator.
- Find \( g'(x) \), the derivative of the denominator.
- Re-evaluate the limit: \( \lim_{x \to c} \frac{f'(x)}{g'(x)} \).
Indeterminate Forms
Indeterminate forms arise in calculus when evaluating limits that do not provide a direct answer, usually resulting from arithmetic operations like multiplication, division, or addition between terms tending towards undefined values, such as infinity or zero. Common indeterminate forms include:
- \( \frac{0}{0} \)
- \( \frac{\infty}{\infty} \)
- \( \infty - \infty \)
- \( 0 \times \infty \)
Limits
A limit in calculus represents the value that a function approaches as the input approaches a certain point. Notated as \( \lim_{x \to c} f(x) \), it is fundamental to calculus, helping to understand behavior of functions at points they might not even be defined. Calculating limits involves:
- Direct substitution, if the function is continuous at the point.
- Factoring, expanding, or simplifying expressions to find a limit, especially when the direct substitution results in an indeterminate form.
- Using special rules or theorems, like L'Hospital's Rule, when the expression simplifies to specific types of indeterminate forms.
Chain Rule
The chain rule is a fundamental technique in calculus used to find the derivative of a composite function. If you have two functions, \( f \) and \( g \), the derivative of \( f(g(x)) \) is expressed as:
- Find the derivative of the outer function \( f \) while keeping the inner function \( g(x) \) as it is: \( f'(g(x)) \).
- Multiply it by the derivative of the inner function \( g(x) \): \( f'(g(x)) \cdot g'(x) \).
- This results in the derivative: \( (f \circ g)'(x) = f'(g(x)) \cdot g'(x) \).
