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Lim(x->0) (e^(x^2)-cosx)/(x^2)...

`Lim_(x->0) (e^(x^2)-cosx)/(x^2)`

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`Lim_(x->0) (e^(x^2)-cosx)/x^2`
It is a `0/0` form.So, we will apply L`'`Hospital rule.
`:. Lim_(x->0) (e^(x^2)-cosx)/x^2 = Lim_(x->0) (e^(x^2)(2x) + sinx)/(2x)`
Again, it is a `0/0` form.So, we will apply L`'`Hospital rule.
`Lim_(x->0) (e^(x^2)(2x) + sinx)/(2x) = Lim_(x->0) (e^(x^2)(2) + 2x(e^(x^2)(2x)) + cosx)/(2)`
Putting `x = 0`, we get,
`= (2+0+1)/2 = 3/2`
`:. Lim_(x->0) (e^(x^2)-cosx)/x^2 = 3/2`
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