# How do I find this limit without L'Hopital's rule? [closed]

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

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

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

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• 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. – Ethan Bolker Oct 3 '18 at 11:20

## 4 Answers

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 $$2$$.

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?

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

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

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

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