# Calculate limit (i(t))/t if i(t)= `int_1^tf(x)dx` f(x)=cosx+2sinx

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### 2 Answers

Assuming that `t->0,` **the limit does not exist. However, each one-sided limit does exist:**

`lim_(t->0+)=-oo` and `lim_(t->0-)=oo` .

If you know a little calculus theory, then you can get this result without any real calculation. First, note that `f(x)` is continuous on `[0,1]`and also, `f(x)>0`on this interval. Thinking in terms of area, this means that `int_0^1 f(x)dx>0,`and so `int_1^0 f(x)dx<0.` ` ` ` ` ` `` `

Also, it's not hard to show that `int_1^t f(x) dx` is continuous, so

`lim_(t->0) int_1^t f(x)dx=int_1^0 f(x)dx<0.` In particular, the value of this integral is finite and nonzero, in other words, bounded. Thus, when dividing by `t` as `t->0,` the entire expression `(i(t))/t` blows up to plus or minus infinity, depending on which direction you approach zero from. If `t>0,` then the numerator approaches a fixed negative number and the denominator goes to zero through positive values, so the limit is negative infinity. If `t<0,` then the numerator approaches the same negative value and the denominator approaches zero through negative values, resulting in a limit of positive infinity.

It looks like a complicated solution when typed out, but the mental steps only take a few seconds to go through.

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Now if `t->oo,` the integral is bounded (essentially because the negative parts keep making up for the positive parts and vice versa-the actual bounds can be found but aren't important here) and the denominator approaches infinity, so the whole thing approaches zero.** In other words**, `lim_(t->oo) (i(t))/t=0.`

The problem does not specify if t goes to 0 or `+-oo` or a specific value.

Considering that t goes to 0, you need to evaluate the following limit, such that:

`lim_(t->0) (i(t))/t`

You need to evaluate the given definite intgeral first, such that:

`int_1^t f(x)dx = int_1^t (cos x + 2sin x)dx`

You need to split the integral into two simpler integrals, using the property of linearity of integral, such that:

`int_1^t (cos x + 2sin x)dx = int_1^t (cos x) dx + int_1^t (2sin x)dx`

`int_1^t (cos x + 2sin x)dx = sin x|_1^t - 2cos x|_1^t`

Using the fundamental theorem of calculus yields:

`int_1^t (cos x + 2sin x)dx = sin t - sin 1 - 2cos t + 2cos 1`

Converting the difference `sin t - sin 1` into a product yields:

`sin t - sin 1 = 2 cos ((t + 1)/2)sin ((t - 1)/2)`

Converting the difference `cos 1 - cos t` into a product yields:

`cos 1 - cos t = 2 sin ((t + 1)/2) sin ((1 - t)/2)`

You need to replace `sin t - sin 1 - 2cos t + 2cos 1` for` i(t)` in limit, such that:

`lim_(t->0) (2 cos ((t + 1)/2)sin ((t - 1)/2) + 4sin ((t + 1)/2) sin ((1 - t)/2))/t =lim_(t->0) (2 cos ((t + 1)/2)sin ((t - 1)/2))/t +lim_(t->0) (4sin ((t + 1)/2) sin ((1 - t)/2))/t`

Replacing 0 for t yields:

`lim_(t->0) (i(t))/t = -oo`

**Hence, evaluating the limit, considering `t->0` , yields **`lim_(t->0) (i(t))/t = -oo.`