- #1

- 2

- 0

## Homework Statement

Lim x-> 0 Sin(x) + Cos(x) + e^-x - 2 / Sin(x) - x

## The Attempt at a Solution

Sin(x) + Cos(x) + e^-x - 2 = 0 and Sin(x) - x = 0

This is indeterminate form?

I used L'Hopitals rule 3 times to get

-Cos(x) + Sin(x) - e^-x / -Cos(x)

The top equals -2 and the bottom equals -1. The limit is 2. Am I understanding this right?

## Homework Statement

Lim x->infinity (sqrt(x^2+1) - 1)^ln(x)

## The Attempt at a Solution

I'm not sure what to do with ln(x). I believe this is the indeterminate form infinity - infinity.

Am I supposed to rewrite this someway?