Limits

[mikagurl]

I am stuck. What is the limit of (Secx-1)/x as x approaches 0. I know to rewrite sec x as 1/cosx. So now that gives me 1/cosx -1 all over x. At this point I tried to copy dot and reciprocal since it was a division problem. Here is where I am stuck. Please help

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[Today123]

Use l'hopitals rule

(sec(x)-1)/x (take derivative of top and bottom) ->>>>> sec(x)tan(x) = tan(x)/cos(x)

when x= 0, tan(0)/cos(0)

so 0/1 = 0

lim x->0 (sec(x)-1)/x =0


http://www.wolframalpha.com/input/?i=lim+x->0+(sec(x)-1)/x+

Hope that make sense, if not there are many youtube vids on l'hopitals rule

 

[HallsofIvy]

A bit tedious but, if you think "L'Hopital's rule" is too heavy for a problem like this, once you have $\displaystyle \frac{\frac{1}{cos(x)}- 1}{x}$ multiply both numerator and denominator by cos(x) to get $\displaystyle \frac{1- cos(x)}{xcos(x)}$.

Now multiply both numerator and denominator by $\displaystyle 1+ cos(x)$: $\displaystyle \dfrac{1- cos^2(x)}{x cos(x)(1+ cos(x))}= \frac{sin^2(x)}{x cos(x)(1+ cos(x))}= \frac{sin(x)}{x}\frac{sin(x)}{cos(x)(1+ cos(x))}$

The first fraction goes to 1 while the second fraction goes to 0.

 

[Quaid]

I tried to copy dot and reciprocal

I don't know what that means. In future posts, please show your steps, so that we may see what you've tried.

If you have not yet learned l'Hôpital's Rule, then study the algebra posted by HallsofIvy.