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Chapter 3: Problem 72

Exponential Limit Evaluate: \(\lim _{x \rightarrow 0}\left(\frac{e^{x^{3}}-1-x^{3}}{\sin ^{6} 2 x}\right)\)

Short Answer

Expert verified
The limit as x approaches 0 of \(\frac{e^{x^{3}}-1-x^{3}}{\sin ^{6} 2x}\) is 0.

Step by step solution

01

Inspecting the Limit Form

We first plug in 0 into the expression. After substitution, \(e^{0^{3}}-1-0^{3}\) evaluates to 0, and \(\sin^{6} (2*0)\) also evaluates to 0. This confirms the limit is in the form 0/0, hence we can apply L'Hopital's rule.
02

Applying L'Hopital's Rule

We are going to differentiate the numerator and the denominator until we get a limit that we can evaluate without getting 0/0. Let's start applying L'Hopital's rule: \(\lim _{x \rightarrow 0}\left(\frac{e^{x^{3}}-1-x^{3}}{\sin ^{6} 2x}\right) = \lim _{x \rightarrow 0}\left(\frac{3xe^{x^{3}}-3x^{2}}{6 \sin ^{5}2x (2\cos(2x))}\right)\). If we substitute x=0 into this new expression, we still get a 0/0 form. So, we apply L'Hopital's rule again to get: \(\lim _{x \rightarrow 0}\left(\frac{3e^{x^{3}}-3x^{2}}{6 \sin ^{5}2x (2\cos(2x))}\right) = \lim _{x \rightarrow 0}\left(\frac{3x^{2}e^{x^{3}}-6x}{-24 \sin ^{4}2x \cos(2x) -12 \sin ^{5}2x \sin(2x)}\right)\)
03

Evaluate the Limit

After differentiating it twice, we have managed to get an evaluation that is not in the 0/0 form. When x=0, the numerator is equal to 0 and the denominator is non-zero, so the limit evaluates to 0.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

L'Hopital's Rule
L'Hopital's Rule is a powerful tool in calculus used to evaluate limits that result in indeterminate forms such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). It states that if the limit \( \lim_{x \to a} \frac{f(x)}{g(x)} \) results in these forms, we can differentiate the numerator \( f(x) \) and the denominator \( g(x) \) and then take the limit again:
  • Differentiate the numerator, \( f'(x) \)
  • Differentiate the denominator, \( g'(x) \)
  • Re-evaluate the limit, \( \lim_{x \to a} \frac{f'(x)}{g'(x)} \)
If the resulting limit is still in an indeterminate form, L'Hopital's Rule can be applied repeatedly until a determinate form is achieved. This makes it an iterative process that can be very helpful when dealing with complex functions that can't be simplified by traditional algebraic methods.
Limit Evaluation
Limit evaluation involves determining the value that a function approaches as the input approaches a particular point. It is foundational in calculus, offering insights into the behavior of functions near specific values. To evaluate a limit:
  • Substitute the value into the function, unless it causes division by zero or another undefined expression.
  • If direct substitution results in an indeterminate form, use techniques like factoring, conjugates, or L'Hopital's Rule to simplify.
This step-by-step process helps in making reasoned decisions about how a function behaves as it nears a certain point. In many cases, particularly for trigonometric or exponential functions, the use of derivatives is key to resolving seemingly complex limits.
Differentiation
Differentiation is the mathematical procedure of finding the derivative of a function. A derivative represents the function's rate of change and is fundamental to calculus. When faced with solving a limit problem using L'Hopital's Rule, differentiation becomes crucial:
  • Derivative of exponential functions like \( e^{x^3} \) involves the chain rule.
  • Trigonometric functions require specific rules, like the derivative of \( \sin(x) \) being \( \cos(x) \).
Understanding how to differentiate correctly can simplify and solve apparent difficulties in limit problems. Differentiation also helps in simplifying expressions to forms where the limits can be easily evaluated.
Trigonometric Limits
Trigonometric limits are calculations involving trigonometric functions as the variable approaches a particular value. For instance, common forms like \( \lim_{x \to 0} \frac{\sin(x)}{x} = 1 \) serve as critical benchmarks. When dealing with trigonometric expressions:
  • Use identities such as \( \sin(2x) = 2 \sin(x)\cos(x) \) to simplify expressions.
  • Recognize standard limits to ease complex trigonometric limit problems.
  • Apply L'Hopital's Rule to transform the indeterminate forms \( \frac{0}{0} \) involving trigonometric functions into solvable forms.
Being well-versed in trigonometric limits and identities allows for straightforward calculations and is especially useful in handling limits involving oscillatory behaviors.

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