Limits and Derivatives — MATHS STD 11 Science — Question

2. 10

Solution:

Given, $ \displaystyle \lim_{\text{x}\rightarrow 1}\frac{2\text{x}^{2}+4\text{x}+4}{2\text{x}-1}:$
 

Substituting x = 1 we get

$ \displaystyle \lim_{\text{x}\rightarrow 1}\frac{2\text{x}^{2}+4\text{x}+4}{2\text{x}-1}=$

$ =\displaystyle \lim_{\text{x}\rightarrow 1}\frac{2\text{(1)}^{2}+4\text{(1)}+4}{2\text{(1)}-1}$

$ =\displaystyle \lim_{\text{x}\rightarrow 1}\frac{2\text{}+4\text{}+4}{2\text{}-1}=10$