MCQ
$\lim _{x \rightarrow 1} \frac{1+\log x-x}{1-2 x+x^2}=$
- A1
- B-1
- C$0$
- ✓$-\frac{1}{2}$
✓
Answer
Correct option: D.
$-\frac{1}{2}$
(D)
Applying L-Hospital's rule, we get
$\lim _{x \rightarrow 1} \frac{1+\log x-x}{1-2 x+x^2}=\lim _{x \rightarrow 1} \frac{\frac{1}{x}-1}{-2+2 x}=\lim _{x \rightarrow 1} \frac{1-x}{2 x(x-1)}$
Again applying L-Hospital's rule, we get
$\lim _{x \rightarrow 1} \frac{-1}{4 x-2}=-\frac{1}{2}$
Applying L-Hospital's rule, we get
$\lim _{x \rightarrow 1} \frac{1+\log x-x}{1-2 x+x^2}=\lim _{x \rightarrow 1} \frac{\frac{1}{x}-1}{-2+2 x}=\lim _{x \rightarrow 1} \frac{1-x}{2 x(x-1)}$
Again applying L-Hospital's rule, we get
$\lim _{x \rightarrow 1} \frac{-1}{4 x-2}=-\frac{1}{2}$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
If $P \left(\frac{\pi}{4}\right)$ is any point on the ellipse $9 x^2+25 y^2=225, S$ and $S ^{\prime}$ are its foci, then $SP . S ^{\prime} P =$________
→→
The circle $x^2+y^2+4 x-4 y+4=0$ touches
→If $\text{x}\neq1$ and $\text{f(x)}=\frac{\text{x}+1}{\text{x}-1}$ is a real function, then $\text{f}(\text{f}(\text{f(2)}))$ is:
→The vertices of a triangle are $(6, 0), (0, 6)$ and $(6, 6)$. The distance between its circumcentre and centroid is :
→The range of $\text{f(x)}=\frac{1}{1-2\cos\text{x}}$ is :
→If $z_1, z_2, z_3$ are complex numbers such that $\left|z_1\right|=\left|z_2\right|=\left|z_3\right|=\left|\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}\right|=1 \text {, }$then $\left|z_1+z_2+z_3\right|$ is
→Let f: R → R be defined as $f(x)=\frac{x^2-x+4}{x^2+x+4}$
Then the range of the function $f (x)$ is
→Then the range of the function $f (x)$ is
Let A = {1, 2, 3} and B = {2, 3, 4} then which of the following is a function from A to B?
If $x_{ n }=\frac{(1-2+3-4+5-6+\ldots)-2 n}{\sqrt{ n ^2+1}+\sqrt{4 n ^2-1}}$, then $\lim _{ n \rightarrow \infty} x_{ n }$ is equal to
→