Solving Limits Using De L'Hospital's Rule

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SUMMARY

The discussion focuses on solving limits using De L'Hospital's Rule, specifically addressing the limit $\lim_{x \to 0} \frac{\sin(x^4)}{5x^4}$. The problem arises from an indeterminate form of $\frac{0}{0}$, allowing the application of De L'Hospital's Rule. The solution confirms that the limit evaluates to $\frac{1}{5}$, leveraging the established fact that $\lim_{u \to 0} \frac{\sin u}{u} = 1$.

PREREQUISITES
  • Understanding of limits in calculus
  • Familiarity with De L'Hospital's Rule
  • Knowledge of trigonometric limits, specifically $\lim_{u \to 0} \frac{\sin u}{u}$
  • Basic integration concepts
NEXT STEPS
  • Study advanced applications of De L'Hospital's Rule in calculus
  • Explore trigonometric limits and their proofs
  • Learn about other indeterminate forms and their resolutions
  • Investigate the implications of limits in real-world scenarios
USEFUL FOR

Students and educators in calculus, mathematicians focusing on limit evaluation, and anyone seeking to deepen their understanding of De L'Hospital's Rule and its applications in solving indeterminate forms.

bahadeen
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can anyone give me a hint on how to solve this probelem View attachment 4167
 

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bahadeen said:
can anyone give me a hint on how to solve this probelem

The limit is equal to $\frac{\int_0^0 \sin(xt^3)dt}{0}= \frac{0}{0}$, so we can use De L'Hospital.We get:$$\lim_{x \to 0} \frac{\sin(x^4)}{5x^4}= \frac{1}{5} \lim_{x \to 0} \frac{\sin(x^4)}{x^4}= \frac{1}{5}$$

since it is known that $\lim_{u \to 0} \frac{\sin u}{u}=1$.
 

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