# A limit problem

I have a problem in limit that made me curious until now. The problem is about the limit and trigonometry. I don't know how to simplify it.. pls help me to solve it. thanks..:)

Note by Leonardo Chandra
5 years, 4 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

## Comments

Sort by:

Top Newest

I assume $$c$$ is a finite value

The key here is L'hôpital's Rule (3 times)

Substitution of $$x=1$$ must give an indeterminate form of $$\frac {0}{0}$$ because denominator yields $$0$$

So the numerator is equals to $$0$$ when $$x=1 \Rightarrow a+b+1=0$$

Apply L'hôpital's Rule and chain rule, the limit becomes

$$\frac {4 a x^{3} + 3 b x^{2}} { \pi (x - 1) \cos (\pi x) + \sin (\pi x) }$$

And it must yield an indeterminate form of $$\frac {0}{0}$$ as well because denominator yields $$0$$ at $$x=1$$, thus numerator must yield $$0$$ as well

So, $$4a+3b=0$$, solve the two equations gives $$a = 3$$, $$b=-4$$

Now substitute those two values into the limit and apply L'hôpital's Rule Rule once again

$$\frac {36 x^{2} - 24 x } { \pi [ -(x-1) \pi \sin (\pi x) + \cos (\pi x) ] + \pi \cos (\pi x) }$$

Substitute $$x=1$$ gives $$c = - \frac {6}{\pi}$$

Hence, $$a = 3, b=-4, c = - \frac {6}{\pi}$$

By the way, I'm new to LaTeX, anyone know how to increase size of the mathematical symbols used? The fractions displayed are too small in standard font size.

- 5 years, 4 months ago

Log in to reply

thanks for the answer, Pi Han..I never consider the using of 0/0 to evaluate the value of numerator, then apply L'hopital more than twice to find the answer. Anyway, thanks for the great answer..I need to learn more again about the L'Hopital rule. :)

- 5 years, 4 months ago

Log in to reply

yeah agreed......with the answer...:)

- 5 years, 4 months ago

Log in to reply

from where did u get this problem?

- 5 years, 4 months ago

Log in to reply

I got this problem from my calculus book. this problem is about limit problems..an introduction to improper form of limit..:)

- 5 years, 4 months ago

Log in to reply

How about We asume that limit when $$x \rightarrow 1$$ is 0/0. Then, calculate it using L'Hopital Theorem.. Sorry, I haven't tried that before..

- 5 years, 4 months ago

Log in to reply

since denominator at x=1 tends to 0 then for the limit to exist ,numerator has to be 0 or else it will form indeterminate form....

0/0 is a basic form in limits....:)

- 5 years, 4 months ago

Log in to reply

thank you very much for explanation :)

- 5 years, 4 months ago

Log in to reply

no problem....

- 5 years, 4 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...