Question:
If the function f(x) satisfies \(\lim_{x\rightarrow 1}\) \(\frac{f(x)-2}{x^2-1}\) =\(\pi\), evaluate \(\lim_{x\rightarrow 1}\) f(x).
If the function f(x) satisfies \(\lim_{x\rightarrow 1}\) \(\frac{f(x)-2}{x^2-1}\) =\(\pi\), evaluate \(\lim_{x\rightarrow 1}\) f(x).
Updated On: Oct 25, 2023
Hide Solution
Verified By Collegedunia
Solution and Explanation
\(\lim_{x\rightarrow 1}\) \(\frac{f(x)-2}{x^2-1}\) =\(\pi\)
\(\Rightarrow\)\(\frac{\lim_{x\rightarrow 1} (f(x)-2)}{\lim_{x\rightarrow 1}(x^2-1)}\) =\(\pi\)
\(\Rightarrow\)\(\lim_{x\rightarrow 1}\) (f(x)-2)=\(\pi\) \(\lim_{x\rightarrow 1}\) (x2-1)
\(\Rightarrow\)\(\lim_{x\rightarrow 1}\) (f(x)-2)=\(\pi\)(12-1)
\(\Rightarrow\)\(\lim_{x\rightarrow 1}\) (f(x)-2)=0
\(\Rightarrow\)\(\lim_{x\rightarrow 1}\) f(x)-\(\lim_{x\rightarrow 1}\)2=0
\(\Rightarrow\)\(\lim_{x\rightarrow 1}\) f(x)- 2=0
∴\(\lim_{x\rightarrow 1}\) f(x) = 2
\(\Rightarrow\)\(\frac{\lim_{x\rightarrow 1} (f(x)-2)}{\lim_{x\rightarrow 1}(x^2-1)}\) =\(\pi\)
\(\Rightarrow\)\(\lim_{x\rightarrow 1}\) (f(x)-2)=\(\pi\) \(\lim_{x\rightarrow 1}\) (x2-1)
\(\Rightarrow\)\(\lim_{x\rightarrow 1}\) (f(x)-2)=\(\pi\)(12-1)
\(\Rightarrow\)\(\lim_{x\rightarrow 1}\) (f(x)-2)=0
\(\Rightarrow\)\(\lim_{x\rightarrow 1}\) f(x)-\(\lim_{x\rightarrow 1}\)2=0
\(\Rightarrow\)\(\lim_{x\rightarrow 1}\) f(x)- 2=0
∴\(\lim_{x\rightarrow 1}\) f(x) = 2
Was this answer helpful?
0
0
Top Questions on Limits
- If \( f(x) = \begin{cases} \frac{(e^x - 1) \log(1 + x)}{x^2} & \text{if } x>0 \\ 1 & \text{if } x = 0 \\ \frac{\cos 4x - \cos bx}{\tan^2 x} & \text{if } x<0 \end{cases} \) is continuous at \( x = 0 \), then \(\sqrt{b^2 - a^2} =\)
- \(\lim_{x \to 0} \frac{x \tan 2x - 2x \tan x}{(1 - \cos 2x)^2} =\)
- Let \([x]\) represent the greatest integer function. If \(\lim_{x \to 0^+} \frac{\cos[x] - \cos(kx - [x])}{x^2} = 5\), then \(k =\)
- Evaluate the limit: \[ \lim_{x \to 0} \frac{(\csc x - \cot x)(e^x - e^{-x})}{\sqrt{3} - \sqrt{2 + \cos x}} \]
- If a real valued function \( f(x) = \begin{cases} (1 + \sin x)^{\csc x} & , -\frac{\pi}{2} < x < 0 \\ a & , x = 0 \\ \frac{e^{2/x} + e^{3/x}}{ae^{2/x} + be^{3/x}} & , 0 < x < \frac{\pi}{2} \end{cases} \) is continuous at \(x = 0\), then \(ab = \)
View More Questions
Questions Asked in CBSE Class XI exam
- Using s, p, d notations, describe the orbital with the following quantum numbers.
- n=1, l=0
- n = 3, l=1
- n = 4, l =2
- n=4, l=3
- CBSE Class XI
- Developments Leading to the Bohr’s Model of Atom
- Write the following as intervals:
(i) {\(x: x ∈ R, -4 < x ≤ 6\)}
(ii) {\(x: x ∈ R, -12 < x < -10\)}
(iii) {\(x: x ∈ R, 0 ≤ x < 7\)}
(iv) {\(x: x ∈ R, 3 ≤ x ≤ 4\)}- CBSE Class XI
- Sets and their Representations
- Reduce the following equations into intercept form and find their intercepts on the axes.
(i) \(3x+2y-12=0\)
(ii) \(4x-3y=6\)
(iii) \(3y+2=0. \)- CBSE Class XI
- Distance of a Point From a Line
- If A = {-1, 1}, find A x A x A.
- CBSE Class XI
- Relations and functions
- Convert the following into basic units:
- 28.7 pm
- 15.15 pm
- 25365 mg
- CBSE Class XI
- Percentage Composition
View More Questions