I tried making t=x-1 but I was left with sin(t(x+1)) which I didn't know what to do with
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Originally Posted by akhayoon I tried making t=x-1 but I was left with sin(t(x+1)) which I didn't know what to do with t = x - 1 we get . We get, Replace and by 1 because that is their limit and we have:
There are many ways to approach these limits. You could let Now, multiply by the conjugate: We can see the familiar sin(t)/t limit. Which equals 1. So you have the limit 1+1=2
sorry,I don't understand how you got nor do I understand how you manipulated the rest of the equation to make those to limits at the end
Here's another attempt. Originally Posted by akhayoon You can see the limit of the first factor exists and it's equal to 1, just make to check it. From here you can easily see that the requested value of the limit is 2.
Originally Posted by akhayoon I tried making t=x-1 but I was left with sin(t(x+1)) which I didn't know what to do with That was a tad simpler I think
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