Limit of $\lim_{x\to1} \frac{\sin{\ln {x}}}{x^5-7x^3+6}$

Question

How does one evaluate this limit without using L'hopital's rule?

$$\lim_{x\to1} \frac{\sin{\ln {x}}}{x^5-7x^3+6}$$

I tried to use the substitution $u=\ln x$ but all I get is an exponential polynomial which can be factored, but it doesn't seem to lead me to a solution

The title says $\lim_{x\rightarrow 3}$ but the body says $\lim_{x\rightarrow 1}$. Which is it? — Jair Taylor, Apr 30 '20 at 06:39
@JairTaylor Considering the limit is trivial at $x=3$, I assume the limit is supposed to go to $1$. — Brevan Ellefsen, Apr 30 '20 at 06:41
@BrevanEllefsen That was my assumption as well. But the OP should clarify. — Jair Taylor, Apr 30 '20 at 06:42

score 4 · Accepted Answer · answered Apr 30 '20 at 06:47

4

Hint: $$ \frac{\sin{\ln {x}}}{x^5-7x^3+6}=\frac{\sin{\ln(x)}}{\ln(x)}\times\frac{\ln(x)}{x-1}\times\dfrac{x-1}{ x^5-7x^3+6}$$ And $$x^5-7x^3+6= (x-1)(x^4+x^3-6x^2-6x-6)$$

answered Apr 30 '20 at 06:47

DINEDINE

6,081

score 3 · Answer 2 · answered Apr 30 '20 at 06:47

3

Hint $$\lim_{x\to1} \frac{\sin{\ln {x}}}{x^5-7x^3+6}=\lim_{x\to1} \frac{\sin{\ln {x}}}{\ln x}\frac{\ln x}{x-1}\frac{x-1}{x^5-7x^3+6}$$

answered Apr 30 '20 at 06:47

Andrei

37,370

Limit of $\lim_{x\to1} \frac{\sin{\ln {x}}}{x^5-7x^3+6}$

2 Answers2