We have an answer to "Evaluate the following in terms of α=limx→0sin⁡xx: limx→0tan2(x)+2xx+x2"

veksetz

Answered question

2022-01-16

Evaluate the following in terms of

α = lim_{x \to 0} \frac{\sin x}{x}

lim_{x \to 0} \frac{\tan^{2} (x) + 2 x}{x + x^{2}}

Lakisha Archer

Beginner2022-01-18Added 39 answers

Given,

lim_{x \to 0} \frac{\tan^{2} (x) + 2 x}{x + x^{2}}

Now taking x common from numerator and denominator,
we get,

lim_{x \to 0} \frac{\tan^{2} (x) + 2 x}{x + x^{2}} = \frac{\tan (x) \frac{\tan x}{x} + 2}{1 + x}

as we know,

lim_{x \to 0} \frac{\tan x}{x} = 1

So,

lim_{x \to 0} \frac{\tan (x) \frac{\tan x}{x} + 2}{1 + x} = \frac{0 \cdot 1 + 2}{1 + 0} = 2

RizerMix

Expert2022-01-20Added 656 answers

You could write

\frac{\tan^{2} x + 2 x}{x + x^{2}} = \frac{\tan^{2} + 2 x}{x (x + 1)} = \frac{\frac{\sin x}{x} \cdot \frac{\sin x}{\cos^{2} x} + 2}{x + 1}

and use limit laws to get the limit

\frac{α \cdot 0 + 2}{1}

alenahelenash

Expert2022-01-24Added 556 answers

Since

lim_{x \to 0} \cos x = 1

we have

lim_{x \to 0} \frac{\tan x}{x} = lim_{x \to 0} \frac{\sin x}{x} = α

, i.e.

\tan x = α x + o (x)

\tan^{2} x = α^{2} x^{2} + o (x^{2})

and

\tan^{2} x + 2 x = 2 x + o (x) = x (2 + o (1))

Then

\frac{\tan^{2} x + 2 x}{x (1 + x)} = \frac{2 + o (1)}{1 + x}

and the limit as

x \to 0

is 2

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