Mathematics Asked by DCdaKING on January 7, 2022

I’m supposed to evaluate the following limit using the cosine of a sum and one of the "special limits" which are ${lim_{xto 0}frac{sin(x)}{x}=1}$ and ${lim_{xto 0}frac{1-cos(x)}{x}=0}$.

The limit is : ${lim_{hto 0}frac{cos(pi + h) + 1}{h}}$

I currently have : ${frac{cos(pi)cos(h) – sin(pi)sin(h) + 1}{h}}$

I’m not sure where to go from here and have been stuck for awhile, help would be very much appreciated.

The hint was given to you:

$$lim_{hto0}frac{cos(pi+h)+1}h=lim_{hto0}frac{cospicos h-sinpisin h+1}h=lim_{hto0}frac{1-cos h}h=0.$$

Answered by user65203 on January 7, 2022

This is actually asking you the derivative of $cos(x)$ at $x=pi$.

We can rewrite the expression as: $$lim_{hrightarrow0}frac{cos(pi+h)-cos(pi)}{pi+h-pi}longrightarrow cos^prime(pi)$$ The derivative of $f(x)$ refers to $lim_{hlongrightarrow0}frac{f(x+h)-f(x)}{x+h-x}$

If you know derivative, you know that $cos^prime(pi)=-sin(pi)=0$

I am not sure if this problem is designed to be solved this way, but it reminds me a problem I did before, and that problem is solved this way.

Hope this is helpful.

Answered by Henry L on January 7, 2022

Since $cos(pi)=-1$ and $sin(pi) =0$ we have: $$frac{cos(pi)cos(h) - sin(pi)sin(h)+1}{h} = frac{-cos(h)+1}{h} = frac{1-cos(h)}{h}$$

Answered by azif00 on January 7, 2022

You have written it correctly. Now because ${sin(pi)=0}$ and ${cos(pi)=-1}$ it simplifies further to

$${frac{cos(pi)cos(h) - sin(pi)sin(h) + 1}{h}=frac{-1times cos(h) - 0times sin(h)+1}{h}=frac{-cos(h) + 1}{h}}$$

can you take it from here?

Answered by Riemann'sPointyNose on January 7, 2022

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