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\[\lim_{x \rightarrow ?1}\frac{ (x^2-1)(\sin(3x-3)) }{\cos(x^3-1)(Tan^2(x^2-x)) }\]
It should say Lim x---> 1
I am not allowed to use L'hopitals rile. rule*
I am not sure how to simplify this. I can't rationalize and I dont see any apparant trignometric identities.
you'll have to make use of this : firstly substitute x-1 --> 0 or y--> where y = x-1 now you must know y->0 sin y /y = 1 and y->0 tan y /y = 1 hope that'll help..
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That didn't make sense...
Where did you get the x-1 from?
x --> 1 right? or in other words, x-1 --> 0 wasnt any miracle !! o.O
Ohh...
haha...
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So I got: |dw:1349974609450:dw|
Now what? @shubhamsrg .
Can I substitute 0 for x-1?
|dw:1349974839369:dw|
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