Evaluate $\lim_{x\to t} \frac{x^2-x-\sqrt{x}+1}{-x^2+x-\sqrt{x}+1}$

Question

$$f(t)=\lim_{x\to t}{\frac{x^2-x-\sqrt{x}+1}{-x^2+x-\sqrt{x}+1}}, \quad t=1,+\infty$$

After drawing a graph, it appears that, $f(1) = - \frac{1}{3} , f(\infty)=-1.$

But I don't know why this is happening. Give me a hint so I can solve the problem.

I'd appreciate it then.

Answer 1

Nothing wrong with l'hopital

$\lim\limits_{x\to 1} \frac{x^2-x-\sqrt{x}+1}{-x^2+x-\sqrt{x}+1}= \lim\limits_{x\to 1}\frac{2x-1-\frac 1{2\sqrt{x}}}{-2x+1-\frac 1{2\sqrt{x}}}=\frac {1-\frac 12}{-1-\frac 12}=-\frac 13$

$\lim\limits_{x\to \infty} \frac{x^2-x-\sqrt{x}+1}{-x^2+x-\sqrt{x}+1}= \lim\limits_{x\to \infty}\frac{2x-1-\frac 1{2\sqrt{x}}}{-2x+1-\frac 1{2\sqrt{x}}}=\lim\limits_{x\to \infty}\frac{2+\frac {\frac 2{\sqrt{x}}}{4x}}{-2+\frac {\frac 2{\sqrt{x}}}{4x}}=\lim\limits_{x\to \infty}\frac{4x\sqrt{x}+ {1}}{-4x\sqrt{x}+1}=\lim\limits_{x\to \infty}\frac{\frac{d(4x\sqrt{x})}{dx}}{-\frac{d(4x\sqrt{x})}{dx}}=-1$

Answer 2

For finite limits $x\to 1$ for example, factoring is usually helpful,

$$\frac{x^2-x-\sqrt{x}+1}{-x^2+x-\sqrt{x}+1}=- \frac{(\sqrt{x}-1)(x\sqrt{x}+x-1)}{(\sqrt{x}-1)(x\sqrt{x}+x+1)}$$ $$=- \frac{x\sqrt{x}+x-1}{x\sqrt{x}+x+1}\to -\frac{1}{3}$$

For limits to $\infty$ divide top and bottom by the highest power, in this case $x^2$.

Answer 3

If you are looking for the limit at t=1, L'Hospital's rule is directly applicable

$$f(t)=\lim_{x\to 1}{\frac{x^2-x-\sqrt{x}+1}{-x^2+x-\sqrt{x}+1}}$$ $$=\lim_{x\to 1}{\frac{2x-1-{\frac{1}{2\sqrt{x}}}}{-2x+1-{\frac{1}{2\sqrt{x}}}}}\to-\frac{1}{3}$$