evaluate-lim-x-3-sinx-2x-1-sinx-x-2-1-

Question Number 174407 by infinityaction last updated on 31/Jul/22

e v a l u a t e

\lim_{x \to - \infty} \frac{3^{\sin x} + 2 x + 1}{\sin x - \sqrt{x^{2} + 1}}

Answered by CElcedricjunior last updated on 31/Jul/22

\lim_{x \to - \infty} \frac{3^{s i n x} + 2 x + 1}{s i n x - \sqrt{x^{2} + 1}}

o r \forall x \in R - 1 ⩽ s i n x ⩽ 1

\Leftrightarrow \frac{1}{3} ⩽ 3^{s i n x} ⩽ 3

\Leftrightarrow \frac{1}{3} + 2 x + 1 ⩽ 3^{s i n x} + 2 x + 1 ⩽ 4 + 2 x

- 1 ⩽ s i n x ⩽ 1

- 1 - \sqrt{x^{2} + 1} ⩽ s i n x - \sqrt{1 + x^{2}} ⩽ 1 - \sqrt{1 + x^{2}}

\Leftrightarrow \frac{\frac{4}{3} + 2 x}{- 1 - \sqrt{1 + x^{2}}} ⩽ \frac{3^{s i n x} + 2 x + 1}{s i n x - \sqrt{1 + x^{2}}} ⩽ \frac{4 + 2 x}{1 - \sqrt{1 + x^{2}}}

\lim_{x \to - \infty} \frac{\frac{4}{3} + 2 x}{- 1 - \sqrt{1 + x^{2}}} = \lim_{x \to - \infty} \frac{\frac{4}{3} + 2 x}{- 1 - ∣ x ∣ \sqrt{1 + \frac{1}{x^{2}}}}

= \lim_{x \to - \infty} \frac{x (\frac{4}{3 x} + 2)}{x (- \frac{1}{x} + \sqrt{1 + \frac{1}{x^{2}}})} c a s q d : {\begin{cases} x - > - \infty \\ ∣ x ∣= - x \end{cases}

= 2

\lim_{x \to - \infty} \frac{4 + 2 x}{1 - \sqrt{1 + x^{2}}} = 2 e n p r o c e d a n t d e

l a m e m e m a n i e r e

d^{'} o \overset{`}{u} \lim_{x \to - \infty} \frac{3^{s i n x} + 2 x + 1}{s i n x - \sqrt{1 + x^{2}}} = 2

D^{'} a p r \overset{`}{e s} l e t h \overset{´}{e o r} \overset{`}{e m e} d e s g e n d a r m e s

\dots \dots \dots l e c \overset{´}{e l} \overset{`}{e b r e} c e d r i c j u n i o r \dots \dots \dots . .

Commented by infinityaction last updated on 01/Aug/22

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