Solveeit Logo
Login

Question

Question: Evaluate the following limit. \(\mathop {\lim }\limits_{x \to 1} \dfrac{{{x^2} + 1}}{{x + 1}}\)...

Evaluate the following limit.
lim⁡x→1x2+1x+1\mathop {\lim }\limits_{x \to 1} \dfrac{{{x^2} + 1}}{{x + 1}}

Explanation

Solution

Hint: Here we need to simplify the limit and then substitute the value of x to evaluate the limit.

Complete step-by-step answer:
Given,
lim⁡x→1x2+1x+1\mathop {\lim }\limits_{x \to 1} \dfrac{{{x^2} + 1}}{{x + 1}}
Add and subtract 2x in the numerator we get
lim⁡x→1x2+1+2x−2xx+1 lim⁡x→1(x+1)2−2xx+1 lim⁡x→1(x+1)2x+1−lim⁡x→12xx+1 lim⁡x→1(x+1)−lim⁡x→12xx+1  \mathop {\lim }\limits_{x \to 1} \dfrac{{{x^2} + 1 + 2x - 2x}}{{x + 1}} \\\ \mathop {\lim }\limits_{x \to 1} \dfrac{{{{(x + 1)}^2} - 2x}}{{x + 1}} \\\ \mathop {\lim }\limits_{x \to 1} \dfrac{{{{(x + 1)}^2}}}{{x + 1}} - \mathop {\lim }\limits_{x \to 1} \dfrac{{2x}}{{x + 1}} \\\ \mathop {\lim }\limits_{x \to 1} (x + 1) - \mathop {\lim }\limits_{x \to 1} \dfrac{{2x}}{{x + 1}} \\\
Now put the value of x we get
(1+1)−2×11+1⇒2−22⇒2−1=1(1 + 1) - \dfrac{{2 \times 1}}{{1 + 1}} \Rightarrow 2 - \dfrac{2}{2} \Rightarrow 2 - 1 = 1
Answer is 1.

Note: We can solve these questions by just putting the value of x but sometimes we have to perform some operation (addition and subtraction) on the denominator or numerator to solve the problem.