Finding Trigonometric Limits: lim,x->0, (5tanx-sinx)/(xcosx)

[Math_Junkie]

Hello all, I need help with solving the following limit:

llim 5tanx - sinx
x->0 xcosx

Thanks in advance.

 

[BigGlenntheHeavy]

Re: Finding Trigonometric Limits

What is the limit of:

\(\displaystyle limit \ \frac{5sec^{2}(x)-cos(x)}{cos(x)-xsin(x)}\)
x-> 0

 

[skeeter]

Re: Finding Trigonometric Limits

Hello all, I need help with solving the following limit:

llim 5tanx - sinx
x->0 xcosx

Thanks in advance.

\(\displaystyle \lim_{x \to 0} \left(\frac{5\tan{x}}{x\cos{x}} - \frac{\sin{x}}{x\cos{x}}\right)\)

\(\displaystyle \lim_{x \to 0} \left(\frac{5\sin{x}}{x\cos^2{x}} - \frac{\sin{x}}{x\cos{x}}\right)\)

\(\displaystyle \lim_{x \to 0} \frac{\sin{x}}{x}\left(\frac{5}{\cos^2{x}} - \frac{1}{\cos{x}}\right) = 1 \cdot \left(\frac{5}{1} - \frac{1}{1}\right) = 4\)

 

[Math_Junkie]

Re: Finding Trigonometric Limits

Thank you so much for the response(s).

Here's a few more I'm having trouble with, I'd appreciate any help as I have a test coming up.
I always hated trig.

lim sin^2(3x)
x->0 4x^2

lim cosx-1
x->0 sinx

lim 1-cosx
x->0 tanx

 

[BigGlenntheHeavy]

Re: Finding Trigonometric Limits

Math_Junkie, are you familiar with the Marquis de l'Hospital?

 

[Math_Junkie]

Re: Finding Trigonometric Limits

Math_Junkie, are you familiar with the Marquis de l'Hospital?

Noup.

 

[Deleted member 4993]

Re: Finding Trigonometric Limits

Thank you so much for the response(s).

Here's a few more I'm having trouble with, I'd appreciate any help as I have a test coming up.
I always hated trig. <<< Then calculus is a wrong place to be

lim sin^2(3x)
x->0 4x^2

\(\displaystyle Lim_{x -> 0}\frac{\sin^2(3x)}{4x^2} \, = \, Lim_{x -> 0}[\frac{\sin^2(3x)}{9x^2} \cdot \frac{9}{4}]\)

lim cosx-1
x->0 sinx

\(\displaystyle Lim_{x -> 0}\frac{\cos(x) - 1}{sin(x)} \, = \, Lim_{x -> 0}[\frac{-2\sin^2(\frac{1}{2}x)}{2\cos(\frac{1}{2}x)\cdot\sin(\frac{1}{2}x)}}]\)

lim 1-cosx
x->0 tanx

Please show us your work, indicating exactly where you are stuck - so that we know where to begin to help you.

 

[galactus]

Re: Finding Trigonometric Limits

\(\displaystyle \lim_{x\to 0}\frac{1-cos(x)}{tan(x)}\)

Don't hate trig. It is very useful.

Anyway, we could multiply top and bottom by 1+cos(x):

\(\displaystyle \lim_{x\to 0}\left(\frac{1-cos(x)}{\frac{sin(x)}{cos(x)}}\cdot\frac{1+cos(x)}{1+cos(x)}\right)\)

\(\displaystyle \lim_{x\to 0}\left(\frac{sin(x)\cdot cos(x)}{cos(x)+1}\right)=\frac{0}{2}=0\)