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

pancha3 2020-11-22 Answered
Compute the following limits if they exist limx0secx1x3
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Expert Answer

Delorenzoz
Answered 2020-11-23 Author has 91 answers
Given:
limx0secx1x3
If the value of limxaf(x)g(x) is in indeterminate form like 00,,0, etc, then limxaf(x)g(x)=limxaf(x)g(x)
Explanation:
Here f(x)=secx1 and g(x)=x3
By plugging the value of the limit x= 0,
f(x)=sec(0)1=11=0
g(x)=(03)=0
So, the indeterminate form is 00
The derivative of f(x) is f(x)=sec(x)tan(x)
The derivative od the is g(x)=3x2
limx0secx1x3=limx0secxtanx3x2
Differentiate f(x) and g(x) with respect to x,
f(x)=sec(x)+2sec3(x) and g(x)=6x
limx0secx1x3=limx0secx+2sec3(x)6x
Take 16 outside and rewrite this as
16limx0(sec(x)+2sec3(x))x1
As x approaches 0 from the left side the above limit goes to negative infinity.
So it is, 16=
Thus, limx0secx1x3=
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