Show that if $\underset{x\to a}{lim}f(x)=L$, then $\underset{x\to a}{lim}cos(f(x))=cos(L)$.

gaby131o
2022-09-26
Answered

Show that if $\underset{x\to a}{lim}f(x)=L$, then $\underset{x\to a}{lim}cos(f(x))=cos(L)$.

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Mackenzie Lutz

Answered 2022-09-27
Author has **13** answers

$$\mathrm{cos}(A)-\mathrm{cos}(B)=-2\mathrm{sin}\left(\frac{A+B}{2}\right)\mathrm{sin}\left(\frac{A-B}{2}\right)$$

Apply this with $A=f(x)$ and $B=L$L. Then

$$|\mathrm{cos}(f(x))-\mathrm{cos}(L)|=2|\mathrm{sin}\left(\frac{f(x)+L}{2}\right)\mathrm{sin}\left(\frac{f(x)-L}{2}\right)|.$$

Now, recall that $|\mathrm{sin}(t)|\le |t|$, so,

$$|\mathrm{cos}(f(x))-\mathrm{cos}(L)|\le \frac{1}{2}|f(x)+L||f(x)-L|.$$

And from here, it is easy to conclude a $\delta -\epsilon $ type proof.

Apply this with $A=f(x)$ and $B=L$L. Then

$$|\mathrm{cos}(f(x))-\mathrm{cos}(L)|=2|\mathrm{sin}\left(\frac{f(x)+L}{2}\right)\mathrm{sin}\left(\frac{f(x)-L}{2}\right)|.$$

Now, recall that $|\mathrm{sin}(t)|\le |t|$, so,

$$|\mathrm{cos}(f(x))-\mathrm{cos}(L)|\le \frac{1}{2}|f(x)+L||f(x)-L|.$$

And from here, it is easy to conclude a $\delta -\epsilon $ type proof.

Ivan Buckley

Answered 2022-09-28
Author has **4** answers

For the sake of completeness,

$$\mathrm{cos}(a+b)=\mathrm{cos}(a)\mathrm{cos}(b)-\mathrm{sin}(a)\mathrm{sin}(b)$$

$$\mathrm{cos}(a-b)=\mathrm{cos}(a)\mathrm{cos}(b)+\mathrm{sin}(a)\mathrm{sin}(b)$$

by substracting

$$\mathrm{cos}(a+b)-\mathrm{cos}(a-b)=-2\mathrm{sin}(a)\mathrm{sin}(b).$$

Now, calling $A=a+b$ and $B=a-b$, gives $a=\frac{A+B}{2}$ and $b=\frac{A-B}{2}$.

$$\mathrm{cos}(a+b)=\mathrm{cos}(a)\mathrm{cos}(b)-\mathrm{sin}(a)\mathrm{sin}(b)$$

$$\mathrm{cos}(a-b)=\mathrm{cos}(a)\mathrm{cos}(b)+\mathrm{sin}(a)\mathrm{sin}(b)$$

by substracting

$$\mathrm{cos}(a+b)-\mathrm{cos}(a-b)=-2\mathrm{sin}(a)\mathrm{sin}(b).$$

Now, calling $A=a+b$ and $B=a-b$, gives $a=\frac{A+B}{2}$ and $b=\frac{A-B}{2}$.

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Given

b.

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Suppose that the dunctions q and r are defined as follows.

$q\left(x\right)={x}^{2}+5$

$r\left(x\right)=\sqrt{x+3}$

Find the following.

$q\circ r\left(1\right)=?$

$(r\circ q)\left(1\right)=?$

Find the following.

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Evaluate the function at given value of the independent variable. Simplify the results.

$f\left(x\right)=\sqrt{x+5}$

$f(x+\mathrm{\u25b3}x)$

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Lets have $y:\mathbb{R}\to {\mathbb{R}}^{2}$ and that $f:{\mathbb{R}}^{2}\to {\mathbb{R}}^{2}$, and lets assume that $f(y(x))$ is given and that $y(x)=y({x}_{0})+{\int}_{{x}_{0}}^{x}f(y(t))dt$

I'm a bit confused how there can be a function of $y(t)$ inside of the function definition for $y(x)$.

I took the example that $y(x)=({x}^{2},x)$ and $f(y,z)=(y+z,y-z)$

$$\Rightarrow f(y(x))=f({x}^{2},x)=({x}^{2}+x,{x}^{2}-x)$$

And now if we follow the definition of $y(x)$ we get:

$$y(x)=y({x}_{0})+{\int}_{{x}_{0}}^{x}f(y(t))dt$$

$$y(x)=y({x}_{0})+{\int}_{{x}_{0}}^{x}({t}^{2}+t,{t}^{2}-t)dt$$

$$\Rightarrow y(x)=y({x}_{0})+(\frac{1}{3}{t}^{3}+\frac{1}{2}{t}^{2},\frac{1}{3}{t}^{3}-\frac{1}{2}{t}^{2}){{\textstyle |}}_{{x}_{0}}^{x}$$

$$\Rightarrow y(x)=(\frac{1}{3}{x}^{3}+\frac{1}{2}{x}^{2}+{C}_{1},\frac{1}{3}{x}^{3}-\frac{1}{2}{x}^{2}+{C}_{2})$$

Where is mistake?

I'm a bit confused how there can be a function of $y(t)$ inside of the function definition for $y(x)$.

I took the example that $y(x)=({x}^{2},x)$ and $f(y,z)=(y+z,y-z)$

$$\Rightarrow f(y(x))=f({x}^{2},x)=({x}^{2}+x,{x}^{2}-x)$$

And now if we follow the definition of $y(x)$ we get:

$$y(x)=y({x}_{0})+{\int}_{{x}_{0}}^{x}f(y(t))dt$$

$$y(x)=y({x}_{0})+{\int}_{{x}_{0}}^{x}({t}^{2}+t,{t}^{2}-t)dt$$

$$\Rightarrow y(x)=y({x}_{0})+(\frac{1}{3}{t}^{3}+\frac{1}{2}{t}^{2},\frac{1}{3}{t}^{3}-\frac{1}{2}{t}^{2}){{\textstyle |}}_{{x}_{0}}^{x}$$

$$\Rightarrow y(x)=(\frac{1}{3}{x}^{3}+\frac{1}{2}{x}^{2}+{C}_{1},\frac{1}{3}{x}^{3}-\frac{1}{2}{x}^{2}+{C}_{2})$$

Where is mistake?

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Function f and g are defined by $f\left(x\right)=\sqrt{({x}^{2}-2x)}$ and $g\left(x\right)=3x+4$ . The composite function is undefined for $x\in ]a;b[$ . Find the value of a and b?

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Now, is this statement correct?

$\mathrm{\forall}f\in F$, $\mathrm{\exists}$$g\in F$ so that $(f\circ g)(1)=2$

This is false because suppose the only output $f(x)$ can produce is

$f(x)=3$ and $f(x)=4$

Now, there is no possibility of $(f\circ g)(1)=2$ because $2$ isn't an output of $f$.

Is that correct to say?

Now, is this statement correct?

$\mathrm{\forall}f\in F$, $\mathrm{\exists}$$g\in F$ so that $(f\circ g)(1)=2$

This is false because suppose the only output $f(x)$ can produce is

$f(x)=3$ and $f(x)=4$

Now, there is no possibility of $(f\circ g)(1)=2$ because $2$ isn't an output of $f$.

Is that correct to say?

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