Question

can you help me find the limit for this equation

Answer 1

use l'hopital's rule, take the derivative of the numerator and denominator

Answer 2

The trick is that they likely expect you to already know (unless this is calc 2 in which you use l'hopitals rule)
\[\lim_{x \rightarrow 0}\frac{ sinx }{ x } = 1\]
So you want to manipulate what you have in order to get that limit, in which you can then replace that with 1 and then plug in x = 0 into the remaining equation.

Answer 3

Do you remember the value of \(\displaystyle\lim_{x\to 0}\frac{\sin x}{x}\)? If so, then note that \[\begin{aligned}\lim_{x\to 0}\frac{\tan x}{3x} &= \frac{1}{3}\lim_{x\to 0}\frac{\tan x}{x} \\ &= \frac{1}{3}\lim_{x\to 0}\frac{\sin x}{x\cos x} \\ &= \frac{1}{3}\lim_{x\to 0}\frac{\sin x}{x}\cdot\lim_{x\to 0}\frac{1}{\cos x} \\ &= \ldots\end{aligned}\]Can you wrap things up from here? :-)

Answer 4

oh i haven't heard of that rule before i dont think that we went over it

Answer 5

Yeah, it's a rule often taught in calc 2 is why. Sometimes it pops up in calc 1, though. But this the sinx/x limit is usually taught, meaning l'hopitals isnt needed.

Answer 6

sinx/x is equal to 1 right?

Answer 7

As long as we're talking about the limit as x goes to 0.

Answer 8

ok so we make the tan into sinx/cosx?

Answer 9

Yep. And then you isolate sinx/x and you can turn that into 1, since that is its limit as x goes to 0. From there, you're allowed to plug in x = 0 to finish off the limit problem.