# Limit with trigonometric functions

## Homework Statement

True or false?

lim Tanx/(1-Cosx) = lim Sec2x/Sinx = Infinity

(limits are as x approaches $$\pi$$ from the left)

## The Attempt at a Solution

I tried just plugging $$\pi$$ into x in the first limit, and I ended up getting 0/2, which exists but is just 0. So I said false. Am I simplifying it too much? because what I did seems to make sense but it looks like that's not what they wanted me to do.

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## Answers and Replies

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Isn't it just 0/2=0?

Mark44
Mentor

The original limit was neither of the indeterminate forms [0/0] or [infinity/infinity], so L'Hopital's Rule doesn't apply here. Your first approach was correct.