a) \(lim_{x\rightarrow2}\left(\sqrt{x+2}+2018\right)=lim_{x\rightarrow2}\left(\sqrt{2+2}+2018\right)=2020\)
b)\(lim_{x\rightarrow+\infty}\dfrac{3.4^n+2^n}{5.4^n+3^n}=lim_{x\rightarrow+\infty}\dfrac{3+\left(\dfrac{2}{4}\right)^n}{5+\left(\dfrac{3}{4}\right)^n}=\dfrac{3+0}{5+0}=\dfrac{3}{5}\)
c) \(lim_{x\rightarrow-3}\dfrac{x^2+4x+3}{x^2-9}=lim_{x\rightarrow-3}\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=lim_{x\rightarrow-3}\dfrac{x+1}{x-3}=\dfrac{-3+1}{-3-3}=\dfrac{1}{3}\)
a) limx→2(√x+2+2018)=√2+2+2018=2020limx→2(x+2+2018)=2+2+2018=2020.
b) limn→+∞3.4n+2n5.4n+3n=limn→+∞3+2n4n5+3n4n=limn→+∞3+(12)n5+(34)n=35limn→+∞3.4n+2n5.4n+3n=limn→+∞3+2n4n5+3n4n=limn→+∞3+(12)n5+(34)n=35.
limx→−3x2+4x+3x2−9=limx→−3(x+1)(x+3)(x−3)(x+3)=limx→−3x+1x−3=−3+1−3−3=13limx→−3x2+4x+3x2−9=limx→−3(x+1)(x+3)(x−3)(x+3)=limx→−3x+1x−3=−3+1−3−3=13.
a) limx→2(√x+2+2018)=√2+2+2018=2020limx→2(x+2+2018)=2+2+2018=2020
b) limn→+∞3.4n+2n5.4n+3n=limn→+∞3+2n4n5+3n4n=limn→+∞3+(12)n5+(34)n=35
limn→+∞3.4n+2n5.4n+3n=limn→+∞3+2n4n5+3n4n=limn→+∞3+(12)n5+(34)n=35
c) limx→−3x2+4x+3x2−9=limx→−3(x+1)(x+3)(x−3)(x+3)=limx→−3x+1x−3=−3+1−3−3=13
a, =\(\sqrt{2+2}\) +2018=2020
b, =limx→+∞\(\dfrac{3+\left(\dfrac{2}{4}\right)^n}{5+\left(\dfrac{3}{4}\right)^n}\)=\(\dfrac{3}{5}\)
c, =limx→-3\(\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)=limx→-3\(\dfrac{x+1}{x-3}\)=\(\dfrac{1}{3}\)
a)
\(=\sqrt{2+2}+2018=2020\)
b)
\(\lim\limits_{\rightarrow+\infty}\dfrac{3.4^n+2^n}{5.4^n+3^n}=\lim\limits_{\rightarrow+\infty}\dfrac{3+\dfrac{2^n}{4^n}}{5+\dfrac{3^n}{4^n}}=\lim\limits_{\rightarrow+\infty}\dfrac{3+\left(\dfrac{1}{2}\right)^n}{5+\left(\dfrac{3}{4}\right)^n}=\dfrac{3}{5}\)
a) limx->2 ( \(\sqrt{x+2}\)+2018)= \(\sqrt{2+2}+2018=2020\)
b) lim x->+∞ \(\dfrac{3.4^n+2^n}{5.4^n+3^n}=lim_{n->+\infty}\dfrac{3+\dfrac{2^n}{4^n}}{5+\dfrac{3^n}{4^n}}=lim_{n->+\infty}\dfrac{3+\left(\dfrac{1}{2}\right)^n}{5+\left(\dfrac{3}{4}\right)^n}=\dfrac{3}{5}\)
c)\(lim_{x->-3}\dfrac{x^2+4x+3}{x^2-9}=lim_{x->-3}\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=lim_{x->-3}\dfrac{x+1}{x-3}=\dfrac{1}{3}\)