Compute the following limits if they exist limx→0sec⁡x−1x3

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Delorenzoz · Accepted Answer

Given:limx→0sec⁡x−1x3If the value of limx→af(x)g(x) is in indeterminate form like 00,∞∞,∞0, etc, then limx→af(x)g(x)=limx→af′(x)g′(x)Explanation:Here f(x)=sec⁡x−1 and g(x)=x3By plugging the value of the limit x= 0,f(x)=sec⁡(0)−1=1−1=0g(x)=(03)=0So, the indeterminate form is 00The derivative of f(x) is f′(x)=sec⁡(x)tan⁡(x)The derivative od the is g′(x)=3x2limx→0sec⁡x−1x3=limx→0sec⁡x⋅tan⁡x3x2Differentiate f(x) and g(x) with respect to x,f″(x)=−sec⁡(x)+2sec3⁡(x) and g″(x)=6xlimx→0sec⁡x−1x3=limx→0−sec⁡x+2sec3⁡(x)6xTake 16 outside and rewrite this as16limx→0(−sec⁡(x)+2sec3⁡(x))x−1As x approaches 0 from the left side the above limit goes to negative infinity.So it is, 16−∞=−∞Thus, limx→0sec⁡x−1x3=−∞

Compute the following limits if they exist lim_{xrightarrow0}frac{sec x-1}{x^3}

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