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How can I know the derivability of this function?

Show that $g'(0)neq lim_{xto 0} g'(x)$L'Hospital's rule problem $lim_{xto 0^+}(x^{x}-1)ln(x)$Question on how to find this particular limit..Limit quotient lawTrigonometric limit: Dividing the Numerator and Denominator by $theta;?$If $lim_{xtoinfty}frac{sin x}{x} = 0$ is zero, why does it not work with L'hospital's way?Does L-Hopital's rule fail for $lim _{x to infty} frac{x+sin x}{x+2 sin x}$?Determine if this two-variable piecewise function is continuousHow to check the differentiability of $f(x) = x|x|$ at $x_0 = 0$?When is this function Differentiable?

-2

$begingroup$

I've been working on this problem:

Study the differentiability of the following function $f$ at $x=0$:
$$
f=begin{cases}
dfrac{cos(3x)e^{3x} - e^x}{ln(1+x)} & text{if} x>0\
2 &text{if} x=0\
bigg(1+3ln(2)ln(1-x)bigg)^{dfrac{-1}{sin(3x)}} &text{if} - pi /6 <x<0
end{cases}
$$

---

Attempt:
$f$ is derivable at $x=0$ if:
$$
lim_{hto0+} [f(0+h) - f(0)]/h
= lim_{hto0-} [f(0+h) - f(0)]/h
$$

Then if we aproach to $0+$ we get:
$$
lim_{hto 0+} ([(cos(3h)e^{3h} - e^h/ ln (1+h)]-2)/h
$$
The limit of the first term: $(cos(3h)e^{3h} - e^h / ln (1+h)$ is $2$

So we got an indetermination 0/0 type

I rewrite the limit as: lim (cos(3h)e^(3h)-e^(h)-2ln(h+1))/ln(h+1)h

Then fixing the denominator to ln(h+1)^(1/h) we get 0/1 so the limit is 0

Correct me if I'm wrong, please.

Then if we aproach to $0-$ we get: lim (1+3ln(2)ln(1-h)^(-1/sin(3h))-2/h

The limit of (1+3ln(2)ln(1-h))^(-1/sin(3h)) is 2

So we get again an indetermination 0/0 type

Using L´Hopital and logaritmic derivation we get:

lim (1/sin^2(3h))(-cos(3h))(3)ln(1+3ln(2)ln(1-h))-(1/sin(3h))(1/(1+3ln(2)ln(1-h))(3ln(2)/(1-h))(-1)^(-1/sin(3h)) as h approaches to 0-

I get another indetermination so...

And at this point I'm stuck.

calculus limits derivatives

share|cite|improve this question

edited Mar 21 at 17:04

Jack

27.7k1782204

asked Mar 20 at 12:23

Isabella Sofia KippesIsabella Sofia Kippes

1

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is the $$f(x)=frac{cos(3x)e^{3x}-e^x}{ln(1+x)}$$
  $endgroup$
  – Dr. Sonnhard Graubner
  Mar 20 at 12:29
  

  
  
  
  


• 
  
  
  

  
  7
  

  
  
  
  
  
  

  
  $begingroup$
  Please use mathjax for the formulas.
  $endgroup$
  – maxmilgram
  Mar 20 at 12:33
  

  
  
  
  

add a comment | 

-2

$begingroup$

I've been working on this problem:

Study the differentiability of the following function $f$ at $x=0$:
$$
f=begin{cases}
dfrac{cos(3x)e^{3x} - e^x}{ln(1+x)} & text{if} x>0\
2 &text{if} x=0\
bigg(1+3ln(2)ln(1-x)bigg)^{dfrac{-1}{sin(3x)}} &text{if} - pi /6 <x<0
end{cases}
$$

---

Attempt:
$f$ is derivable at $x=0$ if:
$$
lim_{hto0+} [f(0+h) - f(0)]/h
= lim_{hto0-} [f(0+h) - f(0)]/h
$$

Then if we aproach to $0+$ we get:
$$
lim_{hto 0+} ([(cos(3h)e^{3h} - e^h/ ln (1+h)]-2)/h
$$
The limit of the first term: $(cos(3h)e^{3h} - e^h / ln (1+h)$ is $2$

So we got an indetermination 0/0 type

I rewrite the limit as: lim (cos(3h)e^(3h)-e^(h)-2ln(h+1))/ln(h+1)h

Then fixing the denominator to ln(h+1)^(1/h) we get 0/1 so the limit is 0

Correct me if I'm wrong, please.

Then if we aproach to $0-$ we get: lim (1+3ln(2)ln(1-h)^(-1/sin(3h))-2/h

The limit of (1+3ln(2)ln(1-h))^(-1/sin(3h)) is 2

So we get again an indetermination 0/0 type

Using L´Hopital and logaritmic derivation we get:

lim (1/sin^2(3h))(-cos(3h))(3)ln(1+3ln(2)ln(1-h))-(1/sin(3h))(1/(1+3ln(2)ln(1-h))(3ln(2)/(1-h))(-1)^(-1/sin(3h)) as h approaches to 0-

I get another indetermination so...

And at this point I'm stuck.

calculus limits derivatives

share|cite|improve this question

edited Mar 21 at 17:04

Jack

27.7k1782204

asked Mar 20 at 12:23

Isabella Sofia KippesIsabella Sofia Kippes

1

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is the $$f(x)=frac{cos(3x)e^{3x}-e^x}{ln(1+x)}$$
  $endgroup$
  – Dr. Sonnhard Graubner
  Mar 20 at 12:29
  

  
  
  
  


• 
  
  
  

  
  7
  

  
  
  
  
  
  

  
  $begingroup$
  Please use mathjax for the formulas.
  $endgroup$
  – maxmilgram
  Mar 20 at 12:33
  

  
  
  
  

add a comment | 

$begingroup$

I've been working on this problem:

Study the differentiability of the following function $f$ at $x=0$:
$$
f=begin{cases}
dfrac{cos(3x)e^{3x} - e^x}{ln(1+x)} & text{if} x>0\
2 &text{if} x=0\
bigg(1+3ln(2)ln(1-x)bigg)^{dfrac{-1}{sin(3x)}} &text{if} - pi /6 <x<0
end{cases}
$$

---

Attempt:
$f$ is derivable at $x=0$ if:
$$
lim_{hto0+} [f(0+h) - f(0)]/h
= lim_{hto0-} [f(0+h) - f(0)]/h
$$

Then if we aproach to $0+$ we get:
$$
lim_{hto 0+} ([(cos(3h)e^{3h} - e^h/ ln (1+h)]-2)/h
$$
The limit of the first term: $(cos(3h)e^{3h} - e^h / ln (1+h)$ is $2$

So we got an indetermination 0/0 type

I rewrite the limit as: lim (cos(3h)e^(3h)-e^(h)-2ln(h+1))/ln(h+1)h

Then fixing the denominator to ln(h+1)^(1/h) we get 0/1 so the limit is 0

Correct me if I'm wrong, please.

Then if we aproach to $0-$ we get: lim (1+3ln(2)ln(1-h)^(-1/sin(3h))-2/h

The limit of (1+3ln(2)ln(1-h))^(-1/sin(3h)) is 2

So we get again an indetermination 0/0 type

Using L´Hopital and logaritmic derivation we get:

lim (1/sin^2(3h))(-cos(3h))(3)ln(1+3ln(2)ln(1-h))-(1/sin(3h))(1/(1+3ln(2)ln(1-h))(3ln(2)/(1-h))(-1)^(-1/sin(3h)) as h approaches to 0-

I get another indetermination so...

And at this point I'm stuck.

calculus limits derivatives

share|cite|improve this question

edited Mar 21 at 17:04

Jack

27.7k1782204

asked Mar 20 at 12:23

Isabella Sofia KippesIsabella Sofia Kippes

1

$endgroup$

share|cite|improve this question

edited Mar 21 at 17:04

Jack

27.7k1782204

asked Mar 20 at 12:23

Isabella Sofia KippesIsabella Sofia Kippes

1

share|cite|improve this question

edited Mar 21 at 17:04

Jack

27.7k1782204

asked Mar 20 at 12:23

Isabella Sofia KippesIsabella Sofia Kippes

1

edited Mar 21 at 17:04

Jack

27.7k1782204

edited Mar 21 at 17:04

Jack

27.7k1782204

asked Mar 20 at 12:23

Isabella Sofia KippesIsabella Sofia Kippes

1

asked Mar 20 at 12:23

Isabella Sofia KippesIsabella Sofia Kippes

1

1 Answer
1

active

oldest

votes

1

$begingroup$

Hint:

$$frac{cos(3x)e^{3x}-e^x}{ln(1+x)}=frac{cos(3x)e^{3x}-e^x}{x}cdot frac{x}{ln(1+x)}.$$

The limits

$$ lim_{x to 0+}frac{cos(3x)e^{3x}-e^x}{x}$$

and

$$ lim_{x to 0+}frac{x}{ln(1+x)}$$

are easy to compute !

share|cite|improve this answer

answered Mar 20 at 12:44

FredFred

48.6k11849

$endgroup$

add a comment | 

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1

$begingroup$

Hint:

$$frac{cos(3x)e^{3x}-e^x}{ln(1+x)}=frac{cos(3x)e^{3x}-e^x}{x}cdot frac{x}{ln(1+x)}.$$

The limits

$$ lim_{x to 0+}frac{cos(3x)e^{3x}-e^x}{x}$$

and

$$ lim_{x to 0+}frac{x}{ln(1+x)}$$

are easy to compute !

share|cite|improve this answer

answered Mar 20 at 12:44

FredFred

48.6k11849

$endgroup$

add a comment | 

1

$begingroup$

Hint:

$$frac{cos(3x)e^{3x}-e^x}{ln(1+x)}=frac{cos(3x)e^{3x}-e^x}{x}cdot frac{x}{ln(1+x)}.$$

The limits

$$ lim_{x to 0+}frac{cos(3x)e^{3x}-e^x}{x}$$

and

$$ lim_{x to 0+}frac{x}{ln(1+x)}$$

are easy to compute !

share|cite|improve this answer

answered Mar 20 at 12:44

FredFred

48.6k11849

$endgroup$

add a comment | 

$begingroup$

Hint:

$$frac{cos(3x)e^{3x}-e^x}{ln(1+x)}=frac{cos(3x)e^{3x}-e^x}{x}cdot frac{x}{ln(1+x)}.$$

The limits

$$ lim_{x to 0+}frac{cos(3x)e^{3x}-e^x}{x}$$

and

$$ lim_{x to 0+}frac{x}{ln(1+x)}$$

are easy to compute !

share|cite|improve this answer

answered Mar 20 at 12:44

FredFred

48.6k11849

$endgroup$

share|cite|improve this answer

answered Mar 20 at 12:44

FredFred

48.6k11849

answered Mar 20 at 12:44

FredFred

48.6k11849

answered Mar 20 at 12:44

FredFred

48.6k11849