# Trig Limit

## Homework Statement

Im not good with latex, but Im having trouble with the limit as x approaches -1 of
[sin(x2+3x+2)]/(x3+1)

## The Attempt at a Solution

Ive tried substituting u = x2+3x+2 but I keep getting -1/4 instead of the correct limit 1/3

by long division, x3+1 = (x2+3x+2)(x-3) + 7x + 7

This yields the limit as u approaches 0 of
[sin(u)]/[u(x-3) + 7x + 7]

after simplifying sinu/u, I get the limit as u approaches 0 of

1/[x - 3 + (7x+7/u)] = u/[u(x-3) + 7x+7]

If I plug in x=-1, the limit is -1/4, not 1/3. What am I doing wrong?

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Have you tried L'Hospital's rule?

cristo
Staff Emeritus
Try using L'Hopital's rule ("type 0/0)"

edit: Damn, beaten to it!

Im not up to it yet, is there any way without derivatives?

As a general rule, you can't really compare different operators together.

For example, each of the following types of operations: polynomial, trigonometric, logarithmic, exponential - if any two appear in the same equation, you can only solve it numerically.

Give me an exact solution for xtan(x)=2, or cos(x) - ln(x) = 0.
you can't.
You need L'Hopital's rule unless I'm missing something obvious.

Dick
Homework Helper
You could try to expand both functions around x=-1 in power series but that needs derivatives as well. Why are you afraid of derivatives and l'Hopital? It's really trivial!

Curious3141
Homework Helper

## Homework Statement

Im not good with latex, but Im having trouble with the limit as x approaches -1 of
[sin(x2+3x+2)]/(x3+1)

## The Attempt at a Solution

Ive tried substituting u = x2+3x+2 but I keep getting -1/4 instead of the correct limit 1/3

by long division, x3+1 = (x2+3x+2)(x-3) + 7x + 7

This yields the limit as u approaches 0 of
[sin(u)]/[u(x-3) + 7x + 7]

after simplifying sinu/u, I get the limit as u approaches 0 of

1/[x - 3 + (7x+7/u)] = u/[u(x-3) + 7x+7]

If I plug in x=-1, the limit is -1/4, not 1/3. What am I doing wrong?
You're on the right track, you don't have to use L'Hopital's.

Your error was in trying to work in two variables when you should work in one variable as far as possible. Since u depends on x, your final expression and limit are not correct.

lim as x-> (-1) of [sin(x2+3x+2)]/(x3+1) = lim as x-> (-1) of [(x2+3x+2)]/(x3+1) since lim(y->0) siny/y = 1

Now,

$$\frac{x^2+3x+2}{x^3+1} = \frac{1}{x-3 + \frac{7(x+1)}{x^2 + 3x+2}}$$

From here on, just continue the long division (sort of like continued fractions).

$$\frac{1}{x-3 + \frac{7(x+1)}{x^2 + 3x+2}} = \frac{1}{x-3 + \frac{7}{x+2}}$$

And the limit of that expression as x -> -1 is clearly 1/(-4+7) = 1/3, as required.

VietDao29
Homework Helper
You're on the right track, you don't have to use L'Hopital's.

Your error was in trying to work in two variables when you should work in one variable as far as possible. Since u depends on x, your final expression and limit are not correct.

lim as x-> (-1) of [sin(x2+3x+2)]/(x3+1) = lim as x-> (-1) of [(x2+3x+2)]/(x3+1) since lim(y->0) siny/y = 1

Now,

$$\frac{x^2+3x+2}{x^3+1} = \frac{1}{x-3 + \frac{7(x+1)}{x^2 + 3x+2}}$$

From here on, just continue the long division (sort of like continued fractions).

$$\frac{1}{x-3 + \frac{7(x+1)}{x^2 + 3x+2}} = \frac{1}{x-3 + \frac{7}{x+2}}$$

And the limit of that expression as x -> -1 is clearly 1/(-4+7) = 1/3, as required.
Err... you seemed to have overcomplicated the problem itself.
It goes like this:
First, you have to change it to some form of:
$$\lim_{\alpha \rightarrow 0} \frac{\sin \alpha}{\alpha} = 1$$
And you know that as x tends to -1, the parameter of the sine function (i.e x2 = 3x + 2) also tends to 0, so you will try to make the denominator looks similar to that parameter, so that you can apply the well-known limit above..
$$\lim_{x \rightarrow -1} \frac{\sin (x ^ 2 + 3x + 2)}{x ^ 3 + 1} = \lim_{x \rightarrow -1} \frac{\sin \left[ (x + 1) (x + 2) \right]}{(x + 1) (x ^ 2 - x + 1)} = \lim_{x \rightarrow -1} \frac{\sin \left[ (x + 1) (x + 2) \right] (x + 2)}{(x + 1) (x + 2) (x ^ 2 - x + 1)}$$
$$= \lim_{x \rightarrow -1} \frac{\sin \left[ (x + 1) (x + 2) \right]}{(x + 1) (x + 2)} \times \lim_{x \rightarrow -1} \frac{x + 2}{x ^ 2 - x + 1} = ...$$
It should be pretty easy from here. Can you go from here? :)

Curious3141
Homework Helper
Err... you seemed to have overcomplicated the problem itself.
It goes like this:
First, you have to change it to some form of:
$$\lim_{\alpha \rightarrow 0} \frac{\sin \alpha}{\alpha} = 1$$
And you know that as x tends to -1, the parameter of the sine function (i.e x2 = 3x + 2) also tends to 0, so you will try to make the denominator looks similar to that parameter, so that you can apply the well-known limit above..
$$\lim_{x \rightarrow -1} \frac{\sin (x ^ 2 + 3x + 2)}{x ^ 3 + 1} = \lim_{x \rightarrow -1} \frac{\sin \left[ (x + 1) (x + 2) \right]}{(x + 1) (x ^ 2 - x + 1)} = \lim_{x \rightarrow -1} \frac{\sin \left[ (x + 1) (x + 2) \right] (x + 2)}{(x + 1) (x + 2) (x ^ 2 - x + 1)}$$
$$= \lim_{x \rarrow -1} \frac{\sin \left[ (x + 1) (x + 2) \right]}{(x + 1) (x + 2)} \times \lim_{x \rarrow -1} \frac{x + 2}{x ^ 2 - x + 1} = ...$$
It should be pretty easy from here. Can you go from here? :)
It's really the same thing, just a different way. I was simply using the fact that $$\lim_{x \rightarrow -1} \sin{(x^2+3x+2)} = \lim_{x \rightarrow -1}(x^2+3x+2)$$ right at the start. Nothing "overcomplicated" in that.

Last edited:
Thanks for the explainations. Im not afraid of LHospitals rule, but we just finished trig limits and are just starting derivatives. Needless to say, we havent gone over it formally in class. The problem was given in a section of the book before LHospitals rule, so I wanted to know how it could be solved without resorting to that

Dick