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Question
For what value of k, the function defined by
f(x) = `{{:((log(1 + 2x)sin^0)/x^2",", "for" x ≠ 0),(k",", "for" x = 0):}`
is continuous at x = 0 ?
Options
2
`1/2`
`π/90`
`90/π`
Solution
`bb(π/90)`
Explanation:
Since,
p(x) = `{{:((log(1 + 2x)sin^0)/x^2",", "for" x ≠ 0),(k",", "for" x = 0):}`
= `{{:((log(1 + 2x)sin (πx)/180)/x^2",", "for" x ≠ 0),(k",", "for" x = 0):}`
As, f(x) is continuous at x = 0
So, LHL = f(0)
∵ LHL = `lim_(x rightarrow 0^-) f(x) = lim_(h rightarrow 0) f(0 - h)`
= `lim_(h rightarrow 0) (log(1 + 2(0 - h)sin π/180^circ (0 - h)))/(0 - h)^2`
= `lim_(h rightarrow 0) (log(1 - 2h){-sin (πh)/180})/h^2`
= `lim_(h rightarrow 0) (-2) (log(1 - 2h))/(-2h) xx lim_(h rightarrow 0) ((-)sin (πh)/180)/((πh)/180) xx π/180`
= `(-2) xx (-1) xx 1 xx π/180` ...`[∵ lim_(x rightarrow 0) log (1 + x)/x = 1 and lim_(x rightarrow 0) sinx/x = 1]`
= `π/90`