1

I have the following function which needs to be found. Obviously this function is fairly straight to find the limit using L'hospital's method (which will give $-5$). However, I need to find the limit without L'hospital's rule. $$\lim_{x \to \pi} \frac{\sin5x}{x-\pi}.$$

I've attempted something like this by taking the x out of the denominator, but I'll still get an indeterminate form. $$\lim_{x \to \pi} \frac{{\sin5x}}{x(1-\frac{\pi}{x})}.$$

$$\lim_{x \to \pi} \frac{5}{1-\frac{\pi}{x}}$$

Xander Henderson
  • 29,772
MineSQL
  • 13
  • 2
    Use $\lim_{y \to 0}\sin(y)/y=1$. – Math Lover Oct 20 '17 at 03:38
  • @MathLover I understand that concept, which gives me the 5 in the numerator, however I can't simplify the function to give a determinant form in the denominator. – MineSQL Oct 20 '17 at 03:42
  • There are a couple of algebraic tricks that you might try. Perhaps consider what happens to $x-\pi$ as $x\to \pi$. – Xander Henderson Oct 20 '17 at 03:43
  • If I'm not mistaken, a number over $x-\pi$ that approaches $\pi$ would NOT converge - as they would go to negative and positive infinity. – MineSQL Oct 20 '17 at 03:45

5 Answers5

5

$$\lim_{x \to \pi} \cdot \frac{\sin(5x)}{x-\pi} = \lim_{x \to \pi} \frac{\sin(5x-5\pi + \pi)}{x-\pi} = \lim_{x \to \pi} \frac{\sin(5x-5\pi)\cos(\pi) + \sin(\pi)\cos(5x-5\pi)}{x-\pi} = \lim_{x \to \pi} \frac{\sin(5x-5\pi)\cdot -1 + 0}{x-\pi}$$

$h = x-\pi$

$$-\lim_{h \to 0} \frac{\sin(5h)}{h} = -5$$

John Lou
  • 2,388
  • 13
  • 16
  • 1
    In second step, use $\sin(x+\pi) = -\sin(x)$ to arrive at the limit $-5$ :) – jonsno Oct 20 '17 at 03:49
  • I'm not really following your mathematics here... The denominator is still in indeterminate form. – MineSQL Oct 20 '17 at 03:50
  • +1 for the answer. @MineSQL do you know lim x->0 sin x / x ? Can you see how John has transformed the earlier question to lim x-> 0 sin x / x ? – TenaliRaman Oct 20 '17 at 03:51
  • 1
    The previous comment was before the edit he just made. Thanks for the clarification @John – MineSQL Oct 20 '17 at 03:56
4

This might be similar to some parts of other answers, but this distills what I think is most important: $$ \begin{align} \lim_{x\to\pi}\frac{\sin(5x)-\sin(5\pi)}{x-\pi} &=\left.\frac{\mathrm{d}}{\mathrm{d}x}\sin(5x)\right|_{\large x=\pi}\\ &=5\cos(5\pi)\\[6pt] &=-5 \end{align} $$

robjohn
  • 345,667
  • I like this answer, but I think that it kind of violates the spirit (if not the letter) of computing the limit without L'Hospital. The (typical) reason to pose these kind of questions is to compute limits while avoiding derivatives of trigonometric functions (which can be circular if not done carefully). – Xander Henderson Oct 20 '17 at 13:29
  • The limit is the derivative, by definition. However you do it, you are finding the derivative. – robjohn Oct 20 '17 at 13:34
  • However, if you prefer, this answer computes the derivative, using geometric means. – robjohn Oct 20 '17 at 13:51
  • I think that I have not communicated my concern appropriately---it is one of pedagogy, not one of correctness. The particular style of question asked here is typically presented to beginning calculus students just after they have seen the geometric argument that you link to, but before they complete the derivation of the derivative of the sine function (as the limit is an essential step of that process). Hence the question (typically) artificially requires the student to compute the limit "directly", without invoking the fact that $\frac{\mathrm{d}}{\mathrm{d}x} \sin(x) = \cos(x)$. – Xander Henderson Oct 20 '17 at 15:31
2

These kind of problems are typically introduced around the time that it is shown that $$ \lim_{t\to 0} \frac{\sin(t)}{t} = 1. $$ The trick is to find a way to make that limit appear. In the current context, the first thing that comes to my mind is making it more explicit that the denominator goes to zero as $x \to \pi$. We can do this by replacing $x-\pi$ with another variable: \begin{align} \lim_{x\to \pi} \frac{\sin(5x)}{x-\pi} &= \lim_{y \to 0} \frac{\sin(5(y+\pi))}{y} && \text{(set $y=x-\pi$)} \end{align} At this point, a little bit of algebraic jiggery-pokery seems necessary to make it a little easier to see what is going on. Remember that the ultimate goal is to get the denominator to match the argument of the sine function. \begin{align} \lim_{y \to 0} \frac{\sin(5(y+\pi))}{y} &= \lim_{y \to 0} \frac{\sin(5y + 5\pi)}{y} \\ &= \lim_{y \to 0} \frac{\sin(5y)\cos(5\pi) + \cos(5y)\sin(5\pi)}{y} && \text{(angle addition formula)}\\ &= \lim_{y \to 0} \frac{-\sin(5y)}{y} && \text{($\cos(5\pi) = -1$, $\sin(5\pi) = 0$)} \\ &= \lim_{y\to 0} \frac{-\sin(5y)}{y} \frac{5}{5} \\ &= -5\lim_{y\to 0} \frac{\sin(5y)}{5y} \\ &= -5. && \left(\text{since $\lim_{t\to 0} \frac{\sin(t)}{t} = 1$}\right) \end{align}

Xander Henderson
  • 29,772
  • This makes a lot more sense to convert to a different variable. Is there a way to keep this in mind? I haven't really came across problems like this before (in my very limited) calculus career. – MineSQL Oct 20 '17 at 03:57
  • @MineSQL If you see $\sin(\cdot)$ in the numerator and a term approaching zero in the denominator, it is likely that you are going to use the fact that $\lim_{t\to 0} \frac{\sin(t)}{t} = 1$. The goal, then, is to find a way to make that limit appear. In this case, my first thought was to make it more clear that the denominator goes to zero, and take a limit with respect to that term. – Xander Henderson Oct 20 '17 at 04:00
1

$\sin(5x-5π+5π)= $

$\sin[(5(x-π) -π)+3×2π] =$

$\sin[5(x-π) -π] = -\sin5(x-π).$

$\lim_{x \rightarrow π}\dfrac{\sin5x}{x-π}=$

$\lim_{x \rightarrow π}( -1)5\dfrac{\sin5(x-π)}{5(x-π)}= -5.$

Peter Szilas
  • 20,344
  • 2
  • 17
  • 28
0

Since $\sin(5\pi)=0$:

$$L=\lim_{x \to \pi} \frac{\sin5x}{x-\pi}=\lim_{x \to \pi} \frac{\sin5x-\sin(5\pi)}{x-\pi}=\lim_{x \to \pi} 2{\cos(5(x+\pi)/2)}\frac {\sin(5(x-\pi)/2)}{x-\pi}$$

$$L=-5\lim_{x \to \pi} \frac {\sin(5(x-\pi)/2)}{\frac {5(x-\pi)}{2}}=-5$$

Note that:

$\lim_{x \to \pi} \frac{\sin5x-\sin(5\pi)}{x-\pi}=5\lim_{x \to \pi} \frac{\sin5x-\sin(5\pi)}{5x-5\pi}$ is just the derivative so it is $5\cos(5x)$ with $x=\pi$

user577215664
  • 40,625