calculus help needed

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Re: calculus help needed

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.