Finding limits of a trig function with a rooted denominator.

Question

I have a final coming up soon and was needing help in understanding a faster method in calculating something like this:

$$\lim_{x \to 0^+} \frac{\sin \sqrt{x}}{\sqrt{x}} $$

I know the answer is 1, but the only method I currently know is to calculate this with the lowest reasonable x value I can input. The thing is if I use 0 the value is undefined, but if I use 1 the value is too far from 0 to correctly approximate, so then I'm left with values smaller than 1 to plug into and attempting to put fractions in a rooted denominator will become a time constraint on the test. This method takes too long and I was wanting to know if there is another method to find the limit.

I don't understand, that does give me the correct value, but it also converts the function into cos(x). When should I utilize this method and when should I not? — LordOfChimichangas, Dec 04 '22 at 03:22
But the peak value starts at a positive value, why wouldn't it be a cosine wave? — LordOfChimichangas, Dec 04 '22 at 03:26
Plot $\cos x$ and $\frac{\sin x}{x}$ on the same graph. If they were the same, they would produce the same picture - they don't. — Sean Roberson, Dec 04 '22 at 03:28
I found a nice proof that uses geometry. https://math.stackexchange.com/questions/75130/how-to-prove-that-lim-limits-x-to0-frac-sin-xx-1 — Doug, Dec 04 '22 at 03:35
I can use L'Hopitals rule to solve for this problem? I never thought about that. — LordOfChimichangas, Dec 04 '22 at 03:38
And I can specifically use L'Hopitals rule because the function produces an undefined result at x=0. That actually explains a lot. — LordOfChimichangas, Dec 04 '22 at 03:42
Now, I have another question. How does L'Hopital's rule in actuality aid in finding the discontinuity for a function? It just seems like a jimmy-rigged method in doing so. I guess what I'm really asking is, why does L'Hopital's rule work so well and not fail? — LordOfChimichangas, Dec 04 '22 at 03:56

score 2 · Answer 1 · answered Dec 04 '22 at 03:27

As mentioned in my comment, we want to force the limit we all learned:

$$\lim_{x \to 0} \frac{\sin x}{x} = 1.$$

To do this, we let $x = y^2$ in the proposed limit. Then as $x \to 0^+,$ we have $y \to 0.$ (You should check this.) Then

$$\lim_{x \to 0^+} \frac{\sin \sqrt{x}}{\sqrt{x}} = \lim_{y \to 0} \frac{\sin y}{y}$$

and we're done.

Finding limits of a trig function with a rooted denominator.

1 Answers1