limit
- Thread starter trivun
- Start date
D
Deleted member 4993
Guest
Please share your work with L'Hopital's rule.
Dr.Peterson
Elite Member
- Joined
- Nov 12, 2017
- Messages
- 16,695
You could also do this without L'Hopital, if you know the limits of sin(x)/x, (1-cos(x))/x^2, and (3^x-1)/x. (The last one I know as the derivative of 3^x.) Just break it up into the product of 3 or 4 fractions.
But how is this Beginning Algebra?
Dr.Peterson's method is very nice. Write out the series expansion of the first two expressions, which allows you to find their limits as [MATH]x \to 0[/MATH] and therefore the limit of their reciprocals. You know the limit of [MATH]\cos x[/MATH] as [MATH]x \to 0[/MATH] and recognise [MATH]\lim \limits_{h \to 0} \tfrac{3^h-3^0}{h}[/MATH] as the derivative of of [MATH]3^x[/MATH] at [MATH]x=0[/MATH] (Differentiate [MATH]3^x[/MATH] and evaluate at 0). A nice solution!
The question is [MATH]\hspace1ex \lim \limits_{x \to 0} \hspace2ex \frac{x}{\sin x} \cdot \frac{x^2}{1-\cos x} \cdot \cos x \cdot \frac{3^x-1}{x}[/MATH]
The question is [MATH]\hspace1ex \lim \limits_{x \to 0} \hspace2ex \frac{x}{\sin x} \cdot \frac{x^2}{1-\cos x} \cdot \cos x \cdot \frac{3^x-1}{x}[/MATH]