A question about evaluating a limit.?

Question

Can anyone solve this?
lim(x->infinity) (x+sinx)/(x-sinx)

I do not know if i can use L'Hospital Rule in this case.

Answer 1

For every x<>0, (x+sin(x))/(x-sin(x)) = (1 +sin(x)/x)/(1 - sin(x)/x). Since sin is bounded, sin(x)/x --> 0 as x --> oo. Therefore, both the numerator and the denominator goes to 1 as x --> oo. It follows that lim (x+sin(x))/(x-sin(x)) =1/1 =1.

L'Hospital rule doesn't work here. Though both the numerator and the denominator of the original limit goes to oo as x -> oo, if we take their derivatives you get (1 + cos(x))/(1 - cos(x)). This is a periodic function and doesn't have a limit when x -->oo. So, L'Hospital provides no information at all.

Answer 2

L'Hospital's rule is unhelpful here, because it says if d(top)/d(bottom) converges, then so does the original fraction. Moreover, they have the same limit. In your case, it doesn't

In your case, sinx stays small and the x's get big. the limit is determined by x/x = 1

Answer 3

lim (3 + 2x)/(2x^3 - x^4) x -> -infinity With fractions which incorporate c/x^n, it makes no distinction no remember if the cut back is going to infinity or damaging infinity; it remains 0. So, fixing this as popular, you divide all words by utilising the optimum power of x. for that reason, that's x^4, so lim (3/x^4 + 2/x^3) / (2/x - a million) x -> -infinity word the cut back. (0 + 0)/(0 - a million) 0/(-a million) that's purely 0 There are in straight forward terms 3 situations (questioning off the right of my head) which you will desire to tension approximately limits going to damaging infinity fairly than infinity are: (a million) Exponentials. cut back as x procedures damaging infinity of e^x is definitely 0, yet while it have been effective infinity the cut back does not exist. lim (e^x) = 0, yet to effective infinity, it does not exist. x -> -infinity (2) arctan(x), or inverse tan(x). because of the fact tan(x) has vertical asymptotes as x = -pi/2 and pi/2, the inverse tan function might have horizontal asymptotes at those factors. meaning lim arctan(x) = -pi/2, yet x -> -infinity lim arctan(x) = pi/2 x -> infinity 2 diverse values at damaging infinity and effective infinity. (3) sq. roots. in case you're asked to discover the cut back of something like lim (sqrt(x^2 - 2) - sqrt(3x^2 + a million)) x -> -infinity Pulling the x^2 out of the muse potential you will get something like this: lim (sqrt(x^2(a million - (2/x^2))) - sqrt(x^2(3 + a million/x^2))) ) x -> -infinity which you will then pull x^2 from the muse as |x|. lim (|x| sqrt((a million - (2/x^2))) - |x| sqrt((3 + a million/x^2))) ) x -> -infinity and you will possibly use the conditional assets that |x| = -x (if x procedures damaging infinity), or |x| = x, if x procedures effective infinity.

Answer 4

(x+sinx)/(x-sinx) = 1 + 2sinx/(x-sinx)
in the limit, sinx is bounded while x-> infinite so the fraction -> 0
lim = 1

Answer 5

Since sin x is bounded between -1 and 1 and x is not bounded, this fraction would tend to 1/1

Answer 6

You don't have to.

(x+sinx)/(x-sinx) = (1+sinx/x)/(1-sinx/x)
Lim sinx/x = 0 when x--> infinity so:
lim (1+sinx/x)/(1-sinx/x) = 1