Qus : 2 1 lim
1 8/3 2 4/3 3 2/3 4 1 Go to Discussion Solution
Quick DL Method Solution
Given:
\lim_{x \to 1} \frac{x^4 - 1}{x - 1}
= \lim_{x \to k} \frac{x^3 - k^2}{x^2 - k^2}
LHS using derivative:
\lim_{x \to 1} \frac{x^4 - 1}{x - 1} = \left.\frac{d}{dx}(x^4)\right|_{x=1} = 4x^3|_{x=1} = 4
RHS using DL logic:
\lim_{x \to k} \frac{x^3 - k^2}{x^2 - k^2}
\approx \frac{3k^2(x - k)}{2k(x - k)} = \frac{3k}{2}
Equating both sides:
\frac{3k}{2} = 4 \Rightarrow k = \frac{8}{3}
\boxed{k = \frac{8}{3}}
Qus : 3 3 Let f(x)=\frac{x^2-1}{|x|-1} . Then the value of lim_{x\to-1} f(x) is
1 -1 2 1 3 2 4
3 Go to Discussion
Solution Qus : 5 4 The value of the limit \lim _{{x}\rightarrow0}\Bigg{(}\frac{{1}^x+{2}^x+{3}^x+{4}^x}{4}{\Bigg{)}}^{1/x} is
1 1 2 {3!}^{1/3!} 3 {3!}^{1/4} 4 {4!}^{1/4} Go to Discussion
Solution Qus : 6 4 Which of the following is NOT true?
1 \lim _{{x}\rightarrow\infty}\frac{x}{{e}^x}=0 2 \lim _{{x}\rightarrow0+}\frac{1}{{xe}^{1/x}}=0 3 \lim _{{x}\rightarrow0+}\frac{\sin x}{{1+2x}}=0 4 \lim _{{x}\rightarrow0+}\frac{\cos x}{{1+2x}}=0 Go to Discussion Qus : 7 4 For a\in R (the set of al real numbers), a \ne 1 , \lim _{{n}\rightarrow\infty}\frac{({1}^a+{2}^a+{\ldots+{n}^a})}{{(n+1)}^{a-1}\lbrack(na+1)(na+b)\ldots(na+n)\rbrack}=\frac{1}{60} . Then one of the value of a is
1 5 2 8 3 -15/2 4 -17/2 Go to Discussion Qus : 8 3 The value of {{Lt}}_{x\rightarrow0}\frac{{e}^x-{e}^{-x}-2x}{1-\cos x} is equal to
1 2 2 1 3 0 4 -1 Go to Discussion Solution
Evaluate:
\lim_{x \to 0} \frac{e^x - e^{-x} - 2x}{1 - \cos x}
Step 1: Apply L'Hôpital's Rule (since it's 0/0):
First derivative:
\frac{e^x + e^{-x} - 2}{\sin x}
Still 0/0 → Apply L'Hôpital's Rule again:
\frac{e^x - e^{-x}}{\cos x}
Now,
\lim_{x \to 0} \frac{1 - 1}{1} = 0
Final Answer:
\boxed{0}
Qus : 9 1 \lim_{x\to \infty} (\frac{x+7}{x+2})^{x+5} equal to
1 e^{5} 2 e^{-5} 3 e^{2} 4 e^{-2} Go to Discussion
Solution Qus : 10 1 Let f(x) be a polynomial of degree four, having extreme value at x = 1 and x = 2. If \lim _{{x}\rightarrow0}[1+\frac{f(x)}{{x}^2}]=3 , then f(2) is
1 0 2 4 3 - 8 4 - 4 Go to Discussion Solution Given it has extremum values at x=1 and x=2
⇒f′(1)=0 and f′(2)=0
Given f(x) is a fourth degree polynomial
Let
f(x)=a{x}^4+b{x}^3+c{x}^2+dx+e Given
\lim _{{x}\rightarrow0}[1+\frac{f(x)}{{x}^2}]=3
\lim _{{x}\rightarrow0}\lbrack1+\frac{a{x}^4+b{x}^3+c{x}^2+\mathrm{d}x+e}{{x}^2}\rbrack=3
\lim _{{x}\rightarrow0}\lbrack1+a{x}^2+bx+c+\frac{d}{x}+\frac{e}{{x}^2}\rbrack=3
For limit to have finite value, value of 'd' and 'e' must be 0
⇒d=0 & e=0
Substituting x=0 in limit
⇒ c+1=3
⇒ c=2
f^{\prime}(x)=4a{x}^3+3b{x}^2+2cx+d
x=1 and x=2 are extreme values,
⇒f^{\prime}(1)=0 and $f^{\prime}(2)=0
⇒ 4a+3b+4=0 and 32a+12b+8=0
By solving these equations
we get, a=\frac{1}{2} and b=-2
So,
f(x)=\frac{x^{4}}{2}-2x^{3}+2x^{2}
⇒f(x)=x^{2}(\frac{x^{2}}{2}-2x+2)
⇒f(2)=0
Qus : 11 4 The value of \lim_{x\to a} \frac{\sqrt{a+2x}-\sqrt{3x}}{\sqrt{3a+x}-2\sqrt{x}}
1 \frac{2}{3} 2 \frac{2}{\sqrt{3}} 3 \frac{3\sqrt{3}}{\sqrt{2}} 4 \frac{2}{3\sqrt{3}} Go to Discussion
Solution Qus : 12 4 1 1/2 2 -1/2 3 - 2 4 2 Go to Discussion Solution Function is the form of
therefore using by L'Hospital rule
=
Again apply L'Hospital Rule,
=
Putting x = 0, we get
Qus : 13 2 Find
1 1 2 limit does not exist 3 infinity 4 None of these Go to Discussion
Solution Qus : 14 3 If f(x)=\lim _{{x}\rightarrow0}\, \frac{{6}^x-{3}^x-{2}^x+1}{\log _e9(1-\cos x)} is a real number then \lim _{{x}\rightarrow0}\, f(x)
1 2 2 3 3 \log _e2 4 \log _e3 Go to Discussion
Solution [{"qus_id":"3388","year":"2018"},{"qus_id":"3909","year":"2019"},{"qus_id":"4306","year":"2017"},{"qus_id":"4315","year":"2017"},{"qus_id":"9449","year":"2020"},{"qus_id":"10665","year":"2021"},{"qus_id":"11130","year":"2022"},{"qus_id":"11136","year":"2022"},{"qus_id":"11509","year":"2023"},{"qus_id":"11523","year":"2023"},{"qus_id":"11550","year":"2023"},{"qus_id":"11624","year":"2024"},{"qus_id":"11656","year":"2024"},{"qus_id":"10196","year":"2015"}]
is equal to 
therefore using by L'Hospital rule
