Re: calculus help needed
riddler wrote:Find the limit. Use l'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. If l'Hospital's Rule doesn't apply, explain why.
a) limit of (x+tanx)/(sinx) as x approaches 0
Hard to write with a typewriter keyboard, but:
lim (x --> 0) (x + tan x)/(sin x) =
lim (x --> 0) (x/sin x) + lim (x --> 0) (tan x/sin x) =
lim (x --> 0) (x/sin x) + lim (x --> 0) (1/cos x)
The second of these two limits is clearly 1, since cos (0) is 1.
The first is indeterminate since both numerator and denominator are zero at the limit, but differentiating numerator and denominator separately to apply L'Hospital's rule, we obtain:
lim (x --> 0) (x/sin x) = lim (x --> 0) (1/cos x), which is the same quotient as the second limit and also 1.
Thus:
lim (x --> 0) (x + tan x)/(sin x) =
lim (x --> 0) (1/cos x) + lim (x --> 0) (1/cos x) = 1 + 1 = 2.
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