How do I solve $\lim\limits_{y\to0}\frac{(x+y)\sec(x+y)-x\sec x}y$. I've tried it a few times but I don't quite get the right answer. Any help is appreciated. Thanks! :)
Find $\lim\limits_{y\to0}\frac{(x+y)\sec(x+y)-x\sec x}y$
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limits
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0What is $u$? Are you trying to find the limit as $(x,y) goes to the origin? – 2017-02-02
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1can you use L'Hospital – 2017-02-02
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0Hint. $x$ is a constant, and the limit of the numerator tends to zero, so the limit should be zero – 2017-02-02
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2@mannav That statement is false. The indeterminate form does not have $0$ as its limit. – 2017-02-02
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0@Dr.SonnhardGraubner I was wondering who was going to make this (idiotic) suggestion first. Kudos, you win the prize of "Hey, let us apply a rule without understanding what it is about". – 2017-02-02
2 Answers
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HINT:
Write the derivative of $x\sec(x)$ using the limit definition of the derivative.
Can you proceed now?
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$$\dfrac{(x+h)\sec(x+h)-x\sec x}h=x\cdot\dfrac{\sec(x+h)-\sec x}h+\sec(x+h)$$
Now $\dfrac{\sec(x+h)-\sec x}h=\dfrac1{\cos x\cos(x+h)}\cdot\dfrac{\cos x-\cos(x+h)}h$
Now use Prosthaphaeresis Formulas, $\cos x-\cos(x+h)=2\sin\dfrac{2x+h}2\sin\dfrac h2$
and $\lim_{u\to0}\dfrac{\sin u}u=1$
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0At the end, you could use derivative of cosine instead. – 2017-02-03
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0@SimplyBeautifulArt, Wanted to use limit to calculate derivative, not the the other way round. Actually the required answer is as we all know is $$\dfrac{d(x\sec x)}{dx}$$ – 2017-02-03