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# limit question 3(urgent)

can u show me how to do this w/o using calculus? Because this chapter is before the chapter of differentiation, andwe should be able to do them using using the knowledge of limit.

thanks

yes, I've learnt that

### 1 Answer

- 翻雷滾天 風卷殘雲Lv 79 years agoFavorite Answer
1) It is indeterminate form of 0/0. Hence applying the L'Hospital rule, the limit is equal to:

lim (x → 0) [d(ex + e-x - 2)/dx]/[d(x sin 4x)/dx]

= lim (x → 0) (ex - e-x)/(sin 4x + 4x cos 4x)

still indeterminate form of 0/0. Applying the rula again:

= lim (x → 0) (ex + e-x)/(4 cos 4x + 4 cos 4x - 16x sin 4x)

= 1/4

2) lim (x → ∞) 5x+3/(26x/2 √5)

= (53/√5) lim (x → ∞) 5x/(26x/2)

= (53/√5) lim (x → ∞) (25/26)x/2

= 0 since 25/26 < 1

3) lim (x → -∞) (3 + 2x)/(3x + 9)

= (3 + 0)/(0 + 9) since all 2x and 3x vanish as x tends to negative infinity.

= 1/3

4) lim (x → ∞) (3 + 2x)/(3x + 9) = lim (x → ∞) [31-x + (2/3)x]/(1 + 32-x)= 0 since all 31-x and (2/3)x vanish as x tends to positive infinity5) -1/x <= (sin mx)/x <= 1/xHence by squeezing principle:lim (x → ∞) (sin mx)/x = 0

2011-05-23 23:21:21 補充：

I think you are referring to Q1 only.

BTW, let me ask you if you have learnt:

lim (x → ∞) (1 + 1/x)^x = e

this limit ?

2011-05-24 17:08:21 補充：

1) By this approach:

e^x = 1 + x + x^2/2 + x^3/3 + ...

e^-x = 1 - x + x^2/2 - x^3/3 + ...

Hence:

e^x + e^-x - 2 = x^2 + x^4/12 + ...

= x^2 (1 + x^2/12 + ...)

(e^x + e^-x - 2)/(x sin 4x) = x (1 + x^2/12 + ...)/sin 4x

2011-05-24 17:08:25 補充：

Thus:

lim (x → 0) (e^x + e^-x - 2)/(x sin 4x) = lim (x → 0) x (1 + x^2/12 + ...)/sin 4x

= lim (x → 0) (1 + x^2/12 + ...) lim (x → 0) x/sin 4x

= (1/4) lim (x → 0) 4x/sin 4x

= 1/4

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