limit(x*(1+(sin(x)^2),x=infinity);limit(x*((sin(x))^2),x=infinity);limit(x^2*sin(1/x)),x=infinity);limit(sqrt(x^2-2*x)-x,x=infinity);limit((x*sqrt(x-2)-sqrt(x)),x=infinity);limit(sqrt(abs(x))/x,x=infinity);看不懂上面的看这里x趋向无穷时

Question

limit(x*(1+(sin(x)^2),x=infinity);
limit(x*((sin(x))^2),x=infinity);
limit(x^2*sin(1/x)),x=infinity);
limit(sqrt(x^2-2*x)-x,x=infinity);
limit((x*sqrt(x-2)-sqrt(x)),x=infinity);
limit(sqrt(abs(x))/x,x=infinity);
看不懂上面的看这里
x 趋向无穷时,求 极限
x*(1+(sinx)^2)
x*(sin(x))^2
x^2*sin(1/x)
sqrt(x^2-2*x)-x
x*(sqrt(x-2)-sqr(x))
sqrt(abs(x))/x

Answer 1

我懒得写infinity了
x*(1+(sinx)^2) => +infinity
x*(sin(x))^2 => 没有极限
x^2*sin(1/x) => 没有极限
sqrt(x^2-2*x)-x
= sqrt(x)(sqrt(x-2)-sqrt(x))
= -sqrt(x)/(sqrt(x)-sqrt(x-2)
= -1/(1-(sqrt((x-2)/x)))
= -1/(1-(sqrt(1-2/x)))
=> -infinity
x*(sqrt(x-2)-sqrt(x)) 就是上一题乘以sqrt(x),因此也是-infinity
sqrt(abs(x))/x => 1/sqrt(x) => 0