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Questions › DSE › DSE 2014 › Q22
2 marks, −0.666666666666667Mathematics › Limits, continuity and differentiability › Limits and indeterminate formsEasyabout 1 min

DSE 2014 Question 22: Limits, continuity and differentiability

lim⁡x→0x2cos⁡ ⁣(1x)=\lim_{x\to0}x^2\cos\!\left(\tfrac{1}{x}\right)=limx→0​x2cos(x1​)=

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lim_(x→0)x²cos (1/x)= (A) -1 (B) 0 (C) 1 (D) The limit does not exist

Also written as: lim_(x->0)x^2cos (1/x)=

Limits, continuity and differentiability has come up 122 times in 7 exams since 2006.

What this question tests

This question is on Limits and indeterminate forms, part of Limits, continuity and differentiability in Mathematics.

Evaluating limits directly or by removing indeterminate forms.

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Question from the DSE 2014 paper. Solutions are TestStop's own. Not affiliated with or endorsed by any examination authority.