Limits with sine and square roots

[mtayab1994]

Homework Statement

solve the limit: \lim_{x\rightarrow1}\frac{sin(x-1)}{\sqrt{x}-1}

The Attempt at a Solution

Is there a way of how i can solve this without using l'hospital's rule or taylor series?

 

[Ray Vickson]

Homework Statement

solve the limit: \lim_{x\rightarrow1}\frac{sin(x-1)}{\sqrt{x}-1}

The Attempt at a Solution

Is there a way of how i can solve this without using l'hospital's rule or taylor series?

What is wrong with using l'Hospital's rule or Taylor series? Of course, changing to the variable t = \sqrt{x} simplifies the question a lot.

RGV

 

[mtayab1994]

Well i haven't learned l'hospitals rule or taylor's series so i can't use either of them.

 

[micromass]

Multiply numerator and denominator by \sqrt{x}+1.

 

[mtayab1994]

Multiply numerator and denominator by \sqrt{x}+1.

ok i multiplied by √x+1 and i got:

\lim_{x\rightarrow1}\frac{sin(x-1)(\sqrt{x}+1)}{x-1}

now should i cancel the x-1?

 

[micromass]

Do you know the limit

\lim_{x\rightarrow 0}\frac{\sin(x)}{x}

 

[mtayab1994]

Do you know the limit

\lim_{x\rightarrow 0}\frac{\sin(x)}{x}

yes it's 1

 

[micromass]

Now, use that to calculate the limit.

 

[mtayab1994]

alright got it now:

\lim_{x\rightarrow1}\frac{sin(x)}{x}*(\sqrt{x}+1)=2

Is that correct?

 

[Mark44]

alright got it now:

\lim_{x\rightarrow1}\frac{sin(x)}{x}*(\sqrt{x}+1)=2

Is that correct?

No.
\lim_{x\to 1}\frac{sin(x)}{x}=\frac{sin(1)}{1} \neq 1

Split your limit into two limits, and then do some fiddling with the first one so that you can use this limit:
\lim_{y \to 0}\frac{sin(y)}{y} = 1

 

[mtayab1994]

No.
\lim_{x\to 1}\frac{sin(x)}{x}=\frac{sin(1)}{1} \neq 1

Split your limit into two limits, and then do some fiddling with the first one so that you can use this limit:
\lim_{y \to 0}\frac{sin(y)}{y} = 1

We still haven't learned how to split a limit into 2 different limits yet.

 

[Mark44]

Are you sure? It's a very basic property of limits, and one that is presented pretty early.
\lim_{x \to a} f(x)\cdot g(x) = \lim_{x \to a}f(x) \cdot \lim_{x \to a}g(x)

The above is true as long as both limits on the right exist. As I see it, you pretty much need to use this idea in your problem.

 

[mtayab1994]

Are you sure? It's a very basic property of limits, and one that is presented pretty early.
\lim_{x \to a} f(x)\cdot g(x) = \lim_{x \to a}f(x) \cdot \lim_{x \to a}g(x)

The above is true as long as both limits on the right exist. As I see it, you pretty much need to use this idea in your problem.

Yes I know that, but we haven't reached it yet in the lesson hence i can't use it.

 

[micromass]

Yes I know that, but we haven't reached it yet in the lesson hence i can't use it.

Then I'm afraid you can't solve this question.

You can't even solve questions like \lim_{x\rightarrow 1}{x^2} in this case.

 

[SammyS]

ok i multiplied by √x+1 and i got:

\lim_{x\rightarrow1}\frac{sin(x-1)(\sqrt{x}+1)}{x-1}

now should i cancel the x-1?

Do you know the limit

\lim_{x\rightarrow 0}\frac{\sin(x)}{x}

yes it's 1

So, if \displaystyle \lim_{x\rightarrow 0}\frac{\sin(x)}{x}=1\,, then what is \displaystyle \lim_{x-1\rightarrow 0}\frac{\sin(x-1)}{x-1}\,?