Using Continuity to evaluate limit of a trig function

[riri]

Hello!

I was wondering if anyone could expand upon and help me with this as I'm struggling
"Use continuity to evalute \(\displaystyle \lim_{{x}\to{\pi}}\cos(x+\sin(x))\)"

I do remember faintly how to do limits of "normal" numbers, but with trig I did not learn at all so I'm confused. This is same as finding the derivative right? But I need to show it in limit form so
\(\displaystyle \lim_{{x}\to{\pi}}\cos(\pi+\sin(\pi)\)?

or Expand it first?

 

[MarkFL]

We are given to find:

\(\displaystyle L=\lim_{x\to\pi}\left(\cos\left(x+\sin(x)\right)\right)\)

Given a continuous function $f:\mathbb{R} \to \mathbb{R}$, we have:

\(\displaystyle \lim_{x\to a} f(x)=f(\lim_{x\to a} x)\)

We can justify the claim that $f$ is continuous if we can show that $f$ is differentiable at every point in $\mathbb{R}$.

After having done so, can you now rewrite the given limit?

By the way, this is a calculus problem, so I am going to move it to our Calculus forum.

 

[riri]

Thank you :)

So... \(\displaystyle \lim_{{x}\to{\pi}}f(x)? = \cos(\lim_{x\to\pi}x+\sin(x)\) ?

\(\displaystyle x(\lim_{{x}\to{a}}\cos+\sin)\)?

 

[MarkFL]

I edited my post to include how we can justify the claim that the function is continuous. We should address that first.

Also, to get your $\LaTeX$ to display properly, you need to wrap your code in [MATH][/MATH] tags.

 

[riri]

Hmm.. so prove that it's differentiable
So would I draw it or something the graph of cos and x?
Or
I think I'm just getting carried away on random ideas
cosx+cossinx?
I wouldn't use the product rule here because I need to use limits, right?
lim(cosx)+limx+sinx

 

[MarkFL]

We have:

\(\displaystyle f(x)=\cos\left(x+\sin(x)\right)\)

So, to find $f^{\prime}(x)$, we need to begin with the rule for differentiating the cosine function and the chain rule:

\(\displaystyle \frac{d}{dx}(\cos(u(x)))=-\sin(u(x))\frac{du}{dx}\)

What do you find?

 

[riri]

Oh! The chain rule needs to be used here? We didn't learn the chain rule yet:( unless it's f(ax))'=af'(ax), is this right?
So -sinx(x+sinx)x(1+cosx)

? I just watched a quick video which said chain rule as outside derivative x inside derivative and this is what I got

 

[MarkFL]

Oh! The chain rule needs to be used here? We didn't learn the chain rule yet:( unless it's f(ax))'=af'(ax), is this right?
So -sinx(x+sinx)x(1+cosx)

? I just watched a quick video which said chain rule as outside derivative x inside derivative and this is what I got

You are almost correct...you have an extra x there, it would be:

\(\displaystyle f^{\prime}(x)=-\sin(x+\sin(x))(1+\cos(x))\)

Is this defined for all real $x$?

 

[riri]

I'm assuming...yes?
Wait... except 0?

Is this chain rule method you showed me how you evaluate using continuity to evaluate this kind of limit? :)

So here's where I have another question. So I'm not done yet I need to use the limit x-->\pi
so it becomes -sin(pi+sin(pi)(1+cos\pi\)
And then...what would would be my next step?
I wouldn't actually evaluate using the real pi number of 3.14 because we're not given calculators
OH! So I would have to find it on the unit circle?
sin(pi)= 0
cos(pi)=-1
So
-sin(pi)(0)?
-sin(pi)?

Is that the final answer? :)

 

[MarkFL]

I'm assuming...yes?
Wait... except 0?

Is this chain rule method you showed me how you evaluate using continuity to evaluate this kind of limit? :)

The chain rule was used only because it was needed to differentiate $f$.

We have:

\(\displaystyle f^{\prime}(0)=-\sin(0+\sin(0))(1+\cos(0))=-\sin(0+0)(1+\cos(0))=-\sin(0)(1+\cos(0))=-0(1+1)=0\cdot2=0\)

So, it is defined...and in fact it is defined for all real $x$, so we know $f$ is continuous. So, now you may write:

\(\displaystyle L=\cos\left(\lim_{x\to\pi}\left(x+\sin(x)\right)\right)\)

For a continuous function, we may use:

\(\displaystyle \lim_{x\to a}f(x)=f(a)\)

Can you proceed?

 

[MarkFL]

...So here's where I have another question. So I'm not done yet I need to use the limit x-->\pi
so it becomes -sin(pi+sin(pi)(1+cos\pi\)
And then...what would would be my next step?
I wouldn't actually evaluate using the real pi number of 3.14 because we're not given calculators
OH! So I would have to find it on the unit circle?
sin(pi)= 0
cos(pi)=-1
So
-sin(pi)(0)?
-sin(pi)?

Is that the final answer? :)

Sorry, I didn't notice you had edited your post until now. This is how I would finish...we have:

\(\displaystyle L=\cos\left(\lim_{x\to\pi}\left(x+\sin(x)\right)\right)\)

And using \(\displaystyle \lim_{x\to a}f(x)=f(a)\) we then may write:

\(\displaystyle L=\cos\left(\pi+\sin(\pi)\right)=\cos\left(\pi+0\right)=\cos\left(\pi\right)=-1\)