I am trying to find an easy way to compute the limit as x→0 off(x)=1+tan⁡(x)−1+sin⁡(x)x3

Solved: "I am trying to find an easy way..."

rhedynogh0rp

2022-03-30

I am trying to find an easy way to compute the limit as

x \to 0

f (x) = \frac{\sqrt{1 + \tan (x)} - \sqrt{1 + \sin (x)}}{x^{3}}

Mercedes Chang

Beginner2022-03-31Added 15 answers

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

= \frac{2 \sin^{2} (\frac{x}{2})}{x^{2}}

For x small enough , we have:

\tan x > x \sin x

(using the geometric interpretation)
then

\cos (x) < \frac{\sin x}{x} < 1

and since the function

\cos (x)

is a continuous function

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

apply this to above, also recall squeeze theorem, we get what you want.
How to prove

\cos x

is continunous, you may ask.

| \cos x - \cos y | = 2 | \sin (\frac{x + y}{2}) \sin (\frac{x - y}{2}) | \leq 2 | \sin (\frac{x - y}{2}) |

now we only have to prove

lim_{x \to 0} \sin x = 0

for x small enough,

0 < | \sin x | < | x |

then

0 \leq lim_{x \to 0} \sin x | \leq lim_{x \to 0} | x ∣ = 0

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