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Question Number 184903 Answers: 1 Comments: 0

lim_(x→0) ((e^x +e^(−x) −2)/x^2 )=?? let x=2t { ((x→0)),((t→0)) :} lim_(x→0) ((e^x +e^(−x) −2)/x^2 )=lim_(t→0) ((e^(2t) +e^(−2t) −2)/(4t^2 ))=(1/4)lim_(t→0) (((e^t −e^(−t) )/t))^2 =(1/4)lim_(t→0) (((e^t −1)/t)+((e^(−t) −1)/(−t)))^2 =(1/4)(1+1)^2 =1

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{e}^{\mathrm{x}} +\mathrm{e}^{−\mathrm{x}} −\mathrm{2}}{\mathrm{x}^{\mathrm{2}} }=?? \\ $$$$\mathrm{let}\:\mathrm{x}=\mathrm{2t}\:\begin{cases}{\mathrm{x}\rightarrow\mathrm{0}}\\{\mathrm{t}\rightarrow\mathrm{0}}\end{cases} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{e}^{\mathrm{x}} +\mathrm{e}^{−\mathrm{x}} −\mathrm{2}}{\mathrm{x}^{\mathrm{2}} }=\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{e}^{\mathrm{2t}} +\mathrm{e}^{−\mathrm{2t}} −\mathrm{2}}{\mathrm{4t}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{4}}\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{e}^{\mathrm{t}} −\mathrm{e}^{−\mathrm{t}} }{\mathrm{t}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{e}^{\mathrm{t}} −\mathrm{1}}{\mathrm{t}}+\frac{\mathrm{e}^{−\mathrm{t}} −\mathrm{1}}{−\mathrm{t}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{1} \\ $$

Question Number 184901 Answers: 2 Comments: 1

lim_((2/x)→2) ((100+x)/(101x))

$$\underset{\frac{\mathrm{2}}{{x}}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\mathrm{100}+{x}}{\mathrm{101}{x}} \\ $$

Question Number 184893 Answers: 1 Comments: 0

find the set of critical points for: 1: f (x) = (x/(⌊ x ⌋+⌊ −x⌋)) −x 2 : g(x) = (x/(⌊ x ⌋ +⌊ −x ⌋)) +x

$$ \\ $$$$\:\:\:\:{find}\:{the}\:{set}\:{of}\: \\ $$$$\:\:\:\:{critical}\:{points}\:{for}: \\ $$$$\:\:\:\mathrm{1}:\:\:\:{f}\:\left({x}\right)\:=\:\frac{{x}}{\lfloor\:{x}\:\rfloor+\lfloor\:−{x}\rfloor}\:−{x}\:\:\:\: \\ $$$$\:\:\:\:\mathrm{2}\::\:\:{g}\left({x}\right)\:=\:\frac{{x}}{\lfloor\:{x}\:\rfloor\:+\lfloor\:−{x}\:\rfloor}\:+{x} \\ $$

Question Number 184891 Answers: 3 Comments: 0

lim_(x→π) ((x−π)/(sinx))=?

$$\underset{{x}\rightarrow\pi} {\mathrm{lim}}\:\frac{{x}−\pi}{{sinx}}=? \\ $$

Question Number 184883 Answers: 1 Comments: 0

lim_(x→2) ((√(x−2))/(x−2))=?

$$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{\sqrt{{x}−\mathrm{2}}}{{x}−\mathrm{2}}=? \\ $$

Question Number 184882 Answers: 1 Comments: 0

lim_(x→0) ((ln(7x+1))/x)=?

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{ln}\left(\mathrm{7}{x}+\mathrm{1}\right)}{{x}}=? \\ $$

Question Number 184881 Answers: 1 Comments: 0

lim_(x→0) ((((√2)sinx)/x))^((sin(x/2)cos(x/2))/x) =?

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\sqrt{\mathrm{2}}{sinx}}{{x}}\right)^{\frac{{sin}\frac{{x}}{\mathrm{2}}{cos}\frac{{x}}{\mathrm{2}}}{{x}}} =? \\ $$

Question Number 184880 Answers: 1 Comments: 0

lim_(x→(1/π)) ((π^(−1) −x)/((1−(xπ)^2 )/π^2 ))=?

$$\underset{{x}\rightarrow\frac{\mathrm{1}}{\pi}} {\mathrm{lim}}\:\:\frac{\pi^{−\mathrm{1}} −{x}}{\frac{\mathrm{1}−\left({x}\pi\right)^{\mathrm{2}} }{\pi^{\mathrm{2}} }}=? \\ $$

Question Number 184879 Answers: 1 Comments: 0

lim_(x→(1/4)) 3x=(3/4) x=?

$$\underset{{x}\rightarrow\frac{\mathrm{1}}{\mathrm{4}}} {\mathrm{lim}}\:\mathrm{3}{x}=\frac{\mathrm{3}}{\mathrm{4}}\:\:\:\:\:\:\:\:\:\:\:{x}=? \\ $$

Question Number 184878 Answers: 1 Comments: 0

lim_(x→0) ((ln(1+sinx))/(sinx))=? with out H′L Roule

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{ln}\left(\mathrm{1}+{sinx}\right)}{{sinx}}=? \\ $$$${with}\:{out}\:{H}'{L}\:{Roule} \\ $$

Question Number 184877 Answers: 1 Comments: 0

Find the sum of the last ten digits: 5^(23) ∙ 2^(2022) − 2023

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{last}\:\mathrm{ten}\:\mathrm{digits}: \\ $$$$\mathrm{5}^{\mathrm{23}} \:\centerdot\:\mathrm{2}^{\mathrm{2022}} \:−\:\mathrm{2023} \\ $$

Question Number 184875 Answers: 1 Comments: 0

Question Number 184873 Answers: 1 Comments: 0

Find the real number satisfying x=(√(1+(√(1+(√(1+x))))))

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{real}\:\mathrm{number}\:\mathrm{satisfying} \\ $$$$\:\mathrm{x}=\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{x}}}} \\ $$

Question Number 184872 Answers: 0 Comments: 0

An odd function f(x) whose domain is R satisfies f(x)=f(x+2). When x ∈ (0, 1), f(x)=−2x^2 +ax−2. If f has 2023 zeros in [0, 1011]. Then the range of a can be ? A. [−6, −2(√2)] B. [−4, −2(√2)] C. [−8, −6] D. [−6, −4]

$$\mathrm{An}\:\mathrm{odd}\:\mathrm{function}\:{f}\left({x}\right)\:\mathrm{whose}\:\mathrm{domain}\:\mathrm{is}\:\mathbb{R} \\ $$$$\mathrm{satisfies}\:{f}\left({x}\right)={f}\left({x}+\mathrm{2}\right).\:\mathrm{When}\:{x}\:\in\:\left(\mathrm{0},\:\mathrm{1}\right), \\ $$$${f}\left({x}\right)=−\mathrm{2}{x}^{\mathrm{2}} +{ax}−\mathrm{2}. \\ $$$$\mathrm{If}\:{f}\:\mathrm{has}\:\mathrm{2023}\:\mathrm{zeros}\:\mathrm{in}\:\left[\mathrm{0},\:\mathrm{1011}\right].\:\mathrm{Then}\:\mathrm{the} \\ $$$$\mathrm{range}\:\mathrm{of}\:{a}\:\mathrm{can}\:\mathrm{be}\:? \\ $$$$\mathrm{A}.\:\left[−\mathrm{6},\:−\mathrm{2}\sqrt{\mathrm{2}}\right]\:\:\:\:\:\:\:\:\mathrm{B}.\:\left[−\mathrm{4},\:−\mathrm{2}\sqrt{\mathrm{2}}\right] \\ $$$$\mathrm{C}.\:\left[−\mathrm{8},\:−\mathrm{6}\right]\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{D}.\:\left[−\mathrm{6},\:−\mathrm{4}\right] \\ $$

Question Number 184869 Answers: 1 Comments: 0