Question

derive x^n-a^n/x-a = na^n-1

derive sinx/x=1

Answer 1

Let

fn(x)=sinn(x)cosn(x)+sinn(x)fn(x)=sinn⁡(x)cosn⁡(x)+sinn⁡(x)

and

f(x)=limn→∞fn(x)f(x)=limn→∞fn(x)

Since both sinsin and coscos are positive on the domain, we have that |fn(x)|≤1|fn(x)|≤1. (In fact, we have <<)

If that's not clear, replace sinnsinn with aa and cosncosn with bb. aa+baa+b is always in [0,1][0,1] if a,b≥0a,b≥0 as long as aa and bb aren't both 00.

Thus, by the dominated convergence theorem[1]

limn→∞∫π30fn(x)dx=∫π30f(x)dxlimn→∞∫0π3fn(x)dx=∫0π3f(x)dx

So let’s figure out what f(x)f(x) is.

For x<π4x<π4, cos(x)>sin(x)cos⁡(x)>sin⁡(x)

Thus sinn(x)sinn⁡(x) goes to 00 faster than cos(x)ncos⁡(x)n, so

limn→∞sinn(x)cosn(x)+sinn(x)=0limn→∞sinn⁡(x)cosn⁡(x)+sinn⁡(x)=0 for x∈[0,π4]x∈[0,π4]

*see end of answer for further explanation

For x>π4x>π4, cos(x)<sin(x)cos⁡(x)<sin⁡(x)

Thus sinn(x)sinn⁡(x) goes to 00 slower than cos(x)ncos⁡(x)n, so

limn→∞sinn(x)cosn(x)+sinn(x)=1limn→∞sinn⁡(x)cosn⁡(x)+sinn⁡(x)=1 for x∈[π4,π3]x∈[π4,π3]

*see end of answer for further explanation

So f(x)=0f(x)=0 if x<π4x<π4

and f(x)=1f(x)=1 if x>π4x>π4

∫π30f(x)dx=1∗(π3−π4)=π12∫0π3f(x)dx=1∗(π3−π4)=π12

That “further explanation” I promised you:

Let’s look at

limn→∞anbn+anlimn→∞anbn+an

==limn→∞1(b/a)n+1limn→∞1(b/a)n+1

If b>ab>a, the denominator goes to ∞∞ and the whole thing goes to 00. If a>ba>b, the denominator goes to 11 and the whole thing goes to 11.

Footnotes

[1] Dominated convergence theorem - Wikipedia