(a) limit as x goes to 0−&gt;tan⁡x2x (b) limit as x goes to 5−&gt;−1(x−5)2 (c) limit as x goes to 0−&gt;cot2x

Step 1 a) The limit can be evaluated as. limx⇒0tan⁡x2x=00 form use LHospitals

Step 2 b) The limit can be evaluated to be as. limx⇒5−1(x−5)2=−10= not defined as function −1(x−5)2 is not defined at x=5 therefore, limx⇒5−1(x−5)2= does not exist (DNE)

Step 3 c) The desired limit can be evaluated to be as. limx⇒0(cot2⁡x)=∞ (not defined) as function (cot2⁡x) approaches ∞ as x tends to 0 therefore, limx⇒0(cot2⁡x)=∞ = not defined

Step 1 a) The limit can be evaluated as. limx⇒0tan⁡x2x=00 form use LHospitals

Step 2 b) The limit can be evaluated to be as. limx⇒5−1(x−5)2=−10= not defined as function −1(x−5)2 is not defined at x=5 therefore, limx⇒5−1(x−5)2= does not exist (DNE)

Step 3 c) The desired limit can be evaluated to be as. limx⇒0(cot2⁡x)=∞ (not defined) as function (cot2⁡x) approaches ∞ as x tends to 0 therefore, limx⇒0(cot2⁡x)=∞ = not defined

Get the answer to "(a) limit as x goes to 0−>tan⁡x2x (b)..."

osula9a

Answered question

2021-12-31

(a) limit as x goes to

0 - > \frac{\tan x}{2 x}

(b) limit as x goes to

5 - > \frac{- 1}{{(x - 5)}^{2}}

0 - > \cot^{2} x

Answer & Explanation

Joseph Lewis

Beginner2022-01-01Added 43 answers

Step 1
a) The limit can be evaluated as.

lim_{x \Rightarrow 0} \frac{\tan x}{2 x} = \frac{0}{0}

form use LHospitals

mauricio0815sh

Beginner2022-01-02Added 34 answers

Step 2
b) The limit can be evaluated to be as.

lim_{x \Rightarrow 5} \frac{- 1}{{(x - 5)}^{2}} = \frac{- 1}{0} =

not defined
as function

\frac{- 1}{{(x - 5)}^{2}}

is not defined at

x = 5

therefore,

lim_{x \Rightarrow 5} \frac{- 1}{{(x - 5)}^{2}} =

does not exist (DNE)

karton

Expert2022-01-04Added 613 answers

Step 3
c) The desired limit can be evaluated to be as.
$lim_{x \Rightarrow 0} (\cot^{2} x) = \infty$ (not defined)
as function $(\cot^{2} x)$ approaches $\infty$ as x tends to 0
therefore,
$lim_{x \Rightarrow 0} (\cot^{2} x) = \infty$ = not defined

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