How Can I Solve These Calculus Limits Without a Calculator?

[Jules18]

Homework Statement

I need to figure out these limits without using a graphing calculator:

lim_x-->0Sinx/(x+Tanx)

lim_x-->0[sqrt(x+4)-2]/x

lim_x-->1(x⁶-1)/(x¹⁰-1)

The Attempt at a Solution

lim_x-->0Sinx/(x+Tanx)
I really had no idea about this one.
Maybe there's an identity I'm forgetting about.

lim_x-->0[sqrt(x+4)-2]/x
For this one, I multiplied the top and bottom by the conjugate of the top, which was
sqrt(x+4) + 2. But it backfired because I ended up with 0/0 as my limit again which
doesn't exist.

lim_x-->1(x⁶-1)/(x¹⁰-1)
For this one, I used a table of values. I got the right answer, but I'm wondering if there's
any way to do it algebraically.


I would be so thankful for any help at all on these ^_^

~Jules~

 

[Mark44]

Homework Statement

I need to figure out these limits without using a graphing calculator:

lim_x-->0Sinx/(x+Tanx)

lim_x-->0[sqrt(x+4)-2]/x

lim_x-->1(x⁶-1)/(x¹⁰-1)

The Attempt at a Solution

lim_x-->0Sinx/(x+Tanx)
I really had no idea about this one.
Maybe there's an identity I'm forgetting about.

lim_x-->0[sqrt(x+4)-2]/x
For this one, I multiplied the top and bottom by the conjugate of the top, which was
sqrt(x+4) + 2. But it backfired because I ended up with 0/0 as my limit again which
doesn't exist.

lim_x-->1(x⁶-1)/(x¹⁰-1)
For this one, I used a table of values. I got the right answer, but I'm wondering if there's
any way to do it algebraically.


I would be so thankful for any help at all on these ^_^

~Jules~

I don't have any ideas for #1, but here are some tips for 2 and 3.
2. You had the right idea, multiplying by the conjugate. After you multiply, you should end up with only x in the numerator, which cancels with the x in the denominator.
3. I haven't worked this through, but both numerator and denominator have factors of (x - 1). Rewrite both numerator and denominator as (x - 1)* (the rest), cancel the x - 1 factors, and you should be able to evaluate the limit.

 

[JG89]

For the first one notice that $$\frac{sin(x)}{x + tan(x)} = \frac{sin(x)}{x(1 + \frac{tan(x)}{x})}$$.

$$\frac{tan(x)}{x} = \frac{\frac{sin(x)}{cos(x)}}{x} = \frac{sin(x)}{xcos(x)}$$

So $$\frac{sin(x)}{x + tan(x)} = \frac{sin(x)}{x} \left \left \left \frac{1}{1 + \frac{sin(x)}{x} + \frac{1}{cos(x)} }$$.

You know the limit of sin(x)/x as x goes to 0, right? So pass to the limit now, using limit laws.

 

[JG89]

I messed up the last equation in my previous post. It should be:

$$\frac{sin(x)}{x + tan(x)} = \frac{sin(x)}{x} \left \left \left \frac{1}{1 + \frac{sin(x)}{x} \left \left \frac{1}{cos(x)} }$$

 

[Jules18]

Thanks guys

JG89 actually I don't know the limit of Sinx/x as x goes to 0 anymore. I'm sure I did at one time. ... what is it again?

 

[JG89]

It's 1

 

[Jules18]

Oookay. that's smart
it ends up being 1/2, which is the same as the answer key.

 

[w3390]

Also just to add a bit more, I believe you can use L'Hopital's Rule when you have 0/0 or infinity/infinity.

 

[HallsofIvy]

Yes, but the title of this thread is "Finding Limits (w/0 calc)".