I have been trying to find
limx→−2x+2√−x−1−1
without L'Hospital's Rule, but I am stuck.
I tried
- Rationalizationg the denominator
- Factoring out x
But it did not work. Finally, I used L'Hospital's Theorem and I got the answer −2.
Is there any way to evaluate this without this concept?
Answer
Rationalizing the denominator works.
limx→−2x+2√−x−1−1=limx→−2x+2√−x−1−1.√−x−1+1√−x−1+1=limx→−2x+2−x−2.(√−x−1+1)=limx→−2−(√−x−1+1)=−2
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