()=\((array)l()^3-9;<2 210;=2 (3(-2)(^2)();> 2(array). ( ) A. Yes, because the limit exist. B. Yes, because the value and the limit are equal. C. No, because the value and the limit are not equal. D. No, because the value and the limit are equal....
Find the limit: $$\lim_{x\rightarrow\pi/6}\sqrt{\csc^{2}x+5\sqrt{3}\tan x} $$ Is the function continuous at the point being approached? Evaluating limits Given a function {eq}f(x) {/eq}, the limit of the functi...
Is the function continuous at {eq}x=3 {/eq}? Explain. Continuity of a function: This problem involves using the concepts of continuity of a function. A function {eq}\displaystyle f(x) {/eq} is said to be continuous at a poi...
Continuity is often confused with contiguity, "drawing the graph in onego," "no gaps," etc. The author argues in support of using correct notions ofcontinuity as well as that of function, where continuity is not to beconsidered in vacuous situations, such as when the function does not ...
0, if x =0 相关知识点: 试题来源: 解析 At x-1.1+1.1=-1.62(x.1.1)=1.1+1.1=1.1+11nπ8-[m,f(0,|b)|=|q|,|d|=1-1|_1+1/(a_1)+1/(a_2)-1/(1+a)-1.5+1+a_(10)- Hence, the given function is discontinuous at x=0 . 反馈 收藏 ...
解析 F(x,y)= (1+x^2+y^2)(1-x^2-y^2) is a rational function and thus is continuous on its domain \( (x,y)|1-x^2-y^2≠q 0\) =\( (x,y)|x^2+y^2≠q 1\) 结果一 题目 Determine the set of points at which the function is continuous. 答案 is continuous on its ...
Show that the function defined byf(x)=cos(x2)is a continuous function. View Solution Is the function defined byf(x)={x+5ifx≤1x−5ifx>1a continuous function? View Solution Exams IIT JEE NEET UP Board Bihar Board CBSE Free Textbook Solutions ...
Give the values of A and B for the function f(x) to be continuous at both x = 1 and x = 6. f(x) = {Ax - B, x less than or equal to 1 : -30 x 1 less than x less than 6: B x^2...
Answer to: Sketch by hand the graph of a function f that satisfies: (a) f(1) = 2 and (b)limx f(x) = 4 Is the function f(x) continuous at x = 1? By...
√ (y-x^2) is continuous on its domain \( (x,y)∣ y-x^2≥q 0\) =\( (x,y)∣ y≥ x^2\) and ln z is continuous on its domain \( z∣ z>0\) , so the product f(x,y,z)=√ (y-x^2) ln z is continuous for y≥ x^2 and z>0, that is, \( (x,y,z)∣ y...