Posted by **John** on Sunday, September 26, 2010 at 8:22pm.

How would you find the limit of (secx-1)/(x^2) as x goes to 0 algebraically?

- Calculus -
**bobpursley**, Sunday, September 26, 2010 at 8:41pm
change secx -1 to 1/cosx -1 to (1-cosx)/cosx

then

lim (1-cosx)/(cosx)x^2 rationalize the numberator.

lim (1-cosx)(1+cosx)/(1+cosx)cosx x^2

lim (sin^2x)/x^2 * 1/(1+cosx)cosx

lim (sinx/x)lim sinx/x Lim (1/(cosx)(1+cosx)

1*1*1/(1*2)= 1/2

- Calculus -
**MathMate**, Sunday, September 26, 2010 at 8:48pm
Multiply top and bottom by (sec(x)+1) to get

(secx-1)(sec(x)+1)/((x^2)(sec(x)+1)

=(sec²(x)-1)/((x^2)(sec(x)+1)

=(sin(x)/x)²/(cos²(x)(1+sec(x)))

Lim x->0 (sin(x)/x)=1

Lim x->0 cos(x)=1

Lim x->0 sec(x)=1

Therefore

Lim x->0 (secx-1)/(x^2) = 1/2

Alternatively, use l'Hôpital's rule.

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