osula9a

2021-12-31

(a) limit as x goes to $0->\frac{\mathrm{tan}x}{2x}$

(b) limit as x goes to$5->\frac{-1}{{(x-5)}^{2}}$

(c) limit as x goes to$0->{\mathrm{cot}}^{2}x$

(b) limit as x goes to

(c) limit as x goes to

Joseph Lewis

Beginner2022-01-01Added 43 answers

Step 1

a) The limit can be evaluated as.

$\underset{x\Rightarrow 0}{lim}\frac{\mathrm{tan}x}{2x}=\frac{0}{0}$ form use LHospitals

a) The limit can be evaluated as.

mauricio0815sh

Beginner2022-01-02Added 34 answers

Step 2

b) The limit can be evaluated to be as.

$\underset{x\Rightarrow 5}{lim}\frac{-1}{{(x-5)}^{2}}=\frac{-1}{0}=$ not defined

as function$\frac{-1}{{(x-5)}^{2}}$ is not defined at $x=5$ therefore,

$\underset{x\Rightarrow 5}{lim}\frac{-1}{{(x-5)}^{2}}=$ does not exist (DNE)

b) The limit can be evaluated to be as.

as function

karton

Expert2022-01-04Added 613 answers

Step 3

c) The desired limit can be evaluated to be as.

as function

therefore,

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