Chapter 9: Problem 696
\(\lim _{\mathrm{X} \rightarrow 1}\left\\{10\left(1-\mathrm{x}^{10}\right)^{-1}-9\left(1-\mathrm{x}^{9}\right)^{-1}\right\\}=?\) (a) \(0.5\) (b) \(0.05\) (c) 45 (d) \(-45\)
Short Answer
Expert verified
The limit does not exist so none of the provided options are correct.
Step by step solution
01
Rewrite the expression.
Rewrite the given expression in a simplified form, so it'll be easier to evaluate the limit. The expression can be rewritten as:
\[\lim _{x \rightarrow 1} \left\\{\frac{10}{1-x^{10}} - \frac{9}{1-x^{9}}\right\\}\]
02
Find common denominator and simplify the expression.
To further simplify the expression, we'll combine the fractions by finding a common denominator, which in this case is \((1-x^{10})(1-x^9)\). Rewrite the expression with the common denominator:
\[\lim _{x \rightarrow 1} \left\\{\frac{10(1-x^9) - 9(1-x^{10})}{(1-x^{10})(1-x^9)} \right\\}\]
Now, expand the numerator:
\[\lim _{x \rightarrow 1} \left\\{\frac{10 - 10x^9 - 9 +9x^{10}}{(1-x^{10})(1-x^9)} \right\\}\]
Next, simplify the numerator by combining like terms:
\[\lim _{x \rightarrow 1} \left\\{\frac{1 - 10x^9 + 9x^{10}}{(1-x^{10})(1-x^9)} \right\\}\]
03
Factor the numerator and cancel out terms.
The numerator seems to have a common factor that we can factor out and simplify:
\[\lim _{x \rightarrow 1} \left\\{\frac{1x^9(1 - 10 + 9x)}{(1-x^{10})(1-x^9)} \right\\}\]
Make sure to distribute the x^9:
\[\lim _{x \rightarrow 1} \left\\{\frac{x^9(1 - x)}{(1-x^{10})(1-x^9)} \right\\}\]
Now, we can see that the term \((1 - x)\) is a common factor in both the numerator and the denominator. Cancel out this term:
\[\lim _{x \rightarrow 1} \left\\{\frac{x^9}{(1-x^9)} \right\\}\]
04
Evaluate the limit.
Now we can safely evaluate the limit as x approaches 1. Substitute 1 for x:
\[\lim _{x \rightarrow 1} \left\\{\frac{1^9}{1-1^9}\right\\}\]
After substituting 1 for x, we get:
\[\frac{1}{1-1} = \frac{1}{0}\]
However, this is undefined. We have made an error in canceling out the common term in the previous step.
Going back to the step, 3:
\[\lim _{x \rightarrow 1} \left\\{\frac{1x^9(1 - 10 + 9x)}{(1-x^{10})(1-x^9)} \right\\}\]
Let's substitute 1 for x:
\[\lim _{x \rightarrow 1} \left\\{\frac{1(1 - 10 + 9)}{(1-1^{10})(1-1^9)} \right\\}\]
\[= \frac{1(-9 + 9)}{(1-1)(1-1)}\]
\[= \frac{0}{0}\]
Since we have 0/0 form, which is an indeterminate form, we need to use L'Hopital's rule.
Applying L'Hôpital's Rule:
Find the derivative of the numerator and denominator:
\[\lim _{x \rightarrow 1} \left\\{\frac{\frac{d}{dx}(1x^9(1 - 10 + 9x))}{\frac{d}{dx}((1-x^{10})(1-x^9))} \right\\}\]
Compute the derivatives:
\[\lim _{x \rightarrow 1} \left\\{\frac{9x^8(1 - 10 + 9x)}{(-10x^9)(1-x^9) - x^{10}(-9x^8)} \right\\}\]
Now, substitute 1 for x:
\[\frac{9(1^8)(1 - 10 + 9)}{(-10)(1^9)(1-1^9) - 1^{10}(-9)(1^8)}\]
\[= \frac{0}{0}\]
We still have the indeterminate form of 0/0, meaning the limit does not exist.
In conclusion, the limit does not exist and is not one of the answer choices provided.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
L'Hôpital's Rule
In calculus, L'Hôpital's Rule is a helpful method used to solve limits of indeterminate forms, like \( \frac{0}{0} \) and \( \frac{\infty}{\infty} \). When you encounter these forms, L'Hôpital's Rule can be your best friend to find the limit by differentiating the numerator and the denominator.Here's a simple way to think about when to use it:
- If direct substitution in a limit problem leads to \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), these are known as indeterminate forms.
- Apply L'Hôpital's Rule by finding the derivative of the numerator and the derivative of the denominator separately.
- After differentiating, recalculate the limit with the new derivatives.
- Repeat the process if necessary, as long as you keep obtaining an indeterminate form.
Indeterminate Forms
In calculus, an indeterminate form is like a traffic signal at an intersection, indicating that the analysis needs some additional steps. Common indeterminate forms to be aware of are:
- \( \frac{0}{0} \) - occurs when both the numerator and the denominator approach zero.
- \( \frac{\infty}{\infty} \) - happens when both tend to infinity.
- \( 0 \cdot \infty \), \( \infty - \infty \), \( 0^0 \), \( 1^\infty \), and \( \infty^0 \).
Evaluating Limits
Evaluating limits involves finding the value that a function approaches as the input approaches a certain point. This is a fundamental skill in calculus that helps in understanding the behavior of functions.
Here's how you can think about evaluating limits:
- The first step is usually to try substituting the value directly into the function. If this leads to a straightforward answer, that's your limit!
- If direct substitution leads to an indeterminate form, consider simplifying the function through factoring, multiplying by a conjugate, or other algebraic techniques.
- If algebraic simplification isn't feasible, apply L'Hôpital's Rule when applicable (as discussed earlier).
- Look for any specific rules, theorems, or patterns that might apply to the type of limit problem you have.