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How would i have to go about finding this limit without using L'Hopital's rule?

$\lim_{t\to 0}\frac{(t)}{{\sqrt{t+1}-cost}}$

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closed as off-topic by Namaste, Scientifica, Key Flex, Delta-u, Paramanand Singh Oct 3 '18 at 16:59

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  • 1
    $\begingroup$ I usually start by writing out the first few terms of the power series for the numerator and denominator. Then the leading terms often tell the story. $\endgroup$ – Ethan Bolker Oct 3 '18 at 11:20
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You have $$\sqrt{1+t} - \cos(t) = (1 + \frac{1}{2}t + o(t)) - (1 +o(t)) = \frac{t}{2} + o(t)$$

So $$\frac{t}{\sqrt{1+t} - \cos(t)} = \frac{1}{\frac{1}{2} + o(1)}$$

And therefore the limit when $t \rightarrow 0$ is $2$.

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Use standard power series: $$\sqrt{1+t} = 1 + \frac{t}{2} + o(t^4)$$ $$\cos t = 1 - \frac{t^2}{2} + o(t^4)$$

Then, we have: $$\lim_{t \to 0}\frac{t}{\left(1 + t/2 + o(t^4) \right) - \left(1 - t^2/2 + o(t^4) \right)}$$

Can you continue?

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I guess $$\frac{t}{\sqrt{1+t}-\cos t}=\frac{t\sqrt{1+t}+t\cos t}{1+t-\cos^2t}=\frac{\sqrt{1+t}+\cos t}{t\left(\frac{\sin t}t\right)^2+1}\to \left[\frac{\sqrt{1+0}+1}{0\cdot 1^2+1}\right]=2$$

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  • $\begingroup$ It is actually strange. The approach which relies on simple facts is downvoted. I upvoted this (+1) $\endgroup$ – Shashi Oct 3 '18 at 11:31
  • $\begingroup$ (I think it was downvoted because before editing, the answer was wrong) $\endgroup$ – TheSilverDoe Oct 3 '18 at 11:44
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Numerator: 1

Denominator :

$d:= \dfrac{\sqrt{t+1}-\cos t}{t}=$

$\dfrac{\sqrt{t+1}-1 +1 -\cos t}{t}=$

$\dfrac{\sqrt{t+1}-1}{t} - \dfrac{\cos (t+0) -\cos 0}{t}=$

Limit $t \rightarrow 0$:

1) $\lim_{t \rightarrow 0}\dfrac{\sqrt{t+1}-1}{t}=$

$(√)' (1)=(1/2)(1)^{-1/2}=1/2$.

2) $-\lim_{t \rightarrow 0} \dfrac{\cos (t+0)-\cos 0}{t}=$

$- (\cos)'(0) = -(-\sin 0)=0.$

Hence

$\dfrac{1}{\lim_{t \rightarrow 0} d} = 2$.

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